Linear Port-Hamiltonian Systems on Infinite-dimensional Spaces

Operator Theory: Advances and Applications Volume 223 Founded in 1979 by Israel Gohberg Editors: Joseph A. Ball (Black...

Author: Birgit Jacob; H J Zwart

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Operator Theory: Advances and Applications Volume 223 Founded in 1979 by Israel Gohberg

Editors: Joseph A. Ball (Blacksburg, VA, USA) Harry Dym (Rehovot, Israel) Marinus A. Kaashoek (Amsterdam, The Netherlands) Heinz Langer (Vienna, Austria) Christiane Tretter (Bern, Switzerland) Associate Editors: Vadim Adamyan (Odessa, Ukraine) Albrecht Böttcher (Chemnitz, Germany) B. Malcolm Brown (Cardiff, UK) Raul Curto (Iowa, IA, USA) Fritz Gesztesy (Columbia, MO, USA) Pavel Kurasov (Lund, Sweden) Leonid E. Lerer (Haifa, Israel) Vern Paulsen (Houston, TX, USA) Mihai Putinar (Santa Barbara, CA, USA) Leiba Rodman (Williamsburg, VA, USA) Ilya M. Spitkovsky (Williamsburg, VA, USA)

Subseries Linear Operators and Linear Systems Subseries editors: Daniel Alpay (Beer Sheva, Israel) Birgit Jacob (Wuppertal, Germany) André C.M. Ran (Amsterdam, The Netherlands) Subseries Advances in Partial Differential Equations Subseries editors: Bert-Wolfgang Schulze (Potsdam, Germany) Michael Demuth (Clausthal, Germany) Jerome A. Goldstein (Memphis, TN, USA) Nobuyuki Tose (Yokohama, Japan) Ingo Witt (Göttingen, Germany)

Honorary and Advisory Editorial Board: Lewis A. Coburn (Buffalo, NY, USA) Ciprian Foias (College Station, TX, USA) J.William Helton (San Diego, CA, USA) Thomas Kailath (Stanford, CA, USA) Peter Lancaster (Calgary, Canada) Peter D. Lax (New York, NY, USA) Donald Sarason (Berkeley, CA, USA) Bernd Silbermann (Chemnitz, Germany) Harold Widom (Santa Cruz, CA, USA)

Birgit Jacob Hans J. Zwart

Linear Port-Hamiltonian Systems on Infinite-dimensional Spaces

L O L S

Linear Operators and Linear Systems

Birgit Jacob Fachbereich C Bergische Universität Wuppertal Wuppertal, Germany

Hans J. Zwart Department of Applied Mathematics University of Twente Enschede, Netherlands

ISBN 978-3-0348-0398-4 ISBN 978-3-0348-0399-1 (eBook) DOI 10.1007/978-3-0348-0399-1 Springer Basel Heidelberg New York Dordrecht London Library of Congress Control Number: 2012940251 © Springer Basel 2012 This work is subject to copyright. All rights are reserved by the Publisher, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microfilms or in any other physical way, and transmission or information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed. Exempted from this legal reservation are brief excerpts in connection with reviews or scholarly analysis or material supplied specifically for the purpose of being entered and executed on a computer system, for exclusive use by the purchaser of the work. Duplication of this publication or parts thereof is permitted only under the provisions of the Copyright Law of the Publisher’s location, in its current version, and permission for use must always be obtained from Springer. Permissions for use may be obtained through RightsLink at the Copyright Clearance Center. Violations are liable to prosecution under the respective Copyright Law. The use of general descriptive names, registered names, trademarks, service marks, etc. in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. While the advice and information in this book are believed to be true and accurate at the date of publication, neither the authors nor the editors nor the publisher can accept any legal responsibility for any errors or omissions that may be made. The publisher makes no warranty, express or implied, with respect to the material contained herein. Printed on acid-free paper

Springer Basel AG is part of Springer Science+Business Media (www.birkhauser-science.com)

Preface The aim of this book is to give a self-contained introduction to the theory of infinite-dimensional systems theory and its applications to port-Hamiltonian systems. The field of infinite-dimensional systems theory has become a well-established field within mathematics and systems theory. There are basically two approaches to infinite-dimensional linear systems theory: an abstract functional analytical approach and a PDE approach. There are excellent books dealing with infinitedimensional linear systems theory, such as (in alphabetical order) Bensoussan, Da Prato, Delfour and Mitter [6], Curtain and Pritchard [9], Curtain and Zwart [10], Fattorini [17], Luo, Guo and Morgul [40], Lasiecka and Triggiani [34, 35], Lions [37], Lions and Magenes [38], Staffans [51], and Tucsnak and Weiss [54]. Many physical systems can be formulated using a Hamiltonian framework. This class contains ordinary as well as partial differential equations. Each system in this class has a Hamiltonian, generally given by the energy function. In the study of Hamiltonian systems it is usually assumed that the system does not interact with its environment. However, for the purpose of control and for the interconnection of two or more Hamiltonian systems it is essential to take this interaction with the environment into account. This led to the class of port-Hamiltonian systems, see [56, 57]. The Hamiltonian/energy has been used to control a port-Hamiltonian system, see e.g. [4, 7, 21, 43]. For port-Hamiltonian systems described by ordinary differential equations this approach is very successful, see the references mentioned above. Port-Hamiltonian systems described by partial differential equations is a subject of current research, see e.g. [14, 28, 33, 41]. In this book, we combine the abstract functional analytical approach with the more physical approach based on Hamiltonians. For a class of linear infinitedimensional port-Hamiltonian systems we derive easily verifiable conditions for well-posedness and stability. The material of this book has been developed over a series of years. Javier Villegas [58] studied in his PhD-thesis a port-Hamiltonian approach to distributed parameter systems. We are grateful to Javier Villegas that we could include his results into the book. The first setup of the book was written for a graduate course on control of distributed parameter systems for the Dutch Institute of Systems and Control (DISC) in the spring of 2009 which was attended by 25 PhD students. This v

vi

Preface

material was adapted for the CIMPA-UNESCO-Marrakech School on Control and Analysis for PDE in May 2009. In 2010-2011 we were the virtual lecturers of the 14th Internet Seminar on Infinite-dimensional Linear Systems Theory. More than 300 participants attended this virtual course and a wikipage was used to discuss the material and to post typos and comments. For this course we decided to add extra chapters on finite-dimensional systems theory, and to make the material in the later chapters more accessible. We are indebted to the help from many colleagues and friends. We are grateful to the participants of the DISC-course, the CIMPA-UNESCO-Marrakesch School and the 14th Internet Seminar for their useful comments and questions. Large parts of the manuscript have been read by our colleagues Mikael Kurula (Twente) and Christian Wyss (Wuppertal), who made many useful comments for improvements. We gratefully acknowledge the financial support of the German Research Foundation (DFG) in form of a Mercator visiting professorship for the second author. Birgit Jacob and Hans Zwart, November 2011 Wuppertal and Twente

Contents List of Figures 1 Introduction 1.1 Examples . . . . . . . . . 1.2 How to control a system? 1.3 Exercises . . . . . . . . . 1.4 Notes and references . . .

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1 1 6 10 12

2 State Space Representation 2.1 State space models . . . . . . . . . 2.2 Solutions of the state space models 2.3 Port-Hamiltonian systems . . . . . 2.4 Exercises . . . . . . . . . . . . . . 2.5 Notes and references . . . . . . . .

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13 13 19 21 24 25

Systems . . . . . . . . . . . . . . . . . . . .

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27 27 33 36 38

4 Stabilizability of Finite-Dimensional Systems 4.1 Stability and stabilizability . . . . . . . . 4.2 The pole placement problem . . . . . . . . 4.3 Characterization of stabilizability . . . . . 4.4 Stabilization of port-Hamiltonian systems 4.5 Exercises . . . . . . . . . . . . . . . . . . 4.6 Notes and references . . . . . . . . . . . .

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39 39 40 44 47 48 49

5 Strongly Continuous Semigroups 5.1 Strongly continuous semigroups . . . . . . . . . . . . . . . . . . . . 5.2 Infinitesimal generators . . . . . . . . . . . . . . . . . . . . . . . .

51 51 57

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3 Controllability of Finite-Dimensional 3.1 Controllability . . . . . . . . . 3.2 Normal forms . . . . . . . . . . 3.3 Exercises . . . . . . . . . . . . 3.4 Notes and references . . . . . .

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vii

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Contents 5.3 5.4 5.5

Abstract differential equations . . . . . . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Notes and references . . . . . . . . . . . . . . . . . . . . . . . . . .

6 Contraction and Unitary Semigroups 6.1 Contraction semigroups . . . . 6.2 Groups and unitary groups . . 6.3 Exercises . . . . . . . . . . . . 6.4 Notes and references . . . . . .

61 62 63

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65 65 73 75 77

7 Homogeneous Port-Hamiltonian Systems 7.1 Port-Hamiltonian systems . . . . . . . 7.2 Generation of contraction semigroups . 7.3 Technical lemmas . . . . . . . . . . . . 7.4 Exercises . . . . . . . . . . . . . . . . 7.5 Notes and references . . . . . . . . . .

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79 79 84 92 93 96

8 Stability 8.1 Exponential stability . . . . . . . . . . . . . 8.2 Spectral projection and invariant subspaces 8.3 Exercises . . . . . . . . . . . . . . . . . . . 8.4 Notes and references . . . . . . . . . . . . .

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97 . 97 . 101 . 108 . 109

9 Stability of Port-Hamiltonian Systems 9.1 Exponential stability of port-Hamiltonian systems 9.2 An example . . . . . . . . . . . . . . . . . . . . . . 9.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . 9.4 Notes and references . . . . . . . . . . . . . . . . .

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111 111 118 120 122

10 Inhomogeneous Abstract Differential Equations and Stabilization 10.1 The abstract inhomogeneous Cauchy problem . . . . . . . . 10.2 Outputs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.3 Bounded perturbations of C0 -semigroups . . . . . . . . . . . 10.4 Exponential stabilizability . . . . . . . . . . . . . . . . . . . 10.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.6 Notes and references . . . . . . . . . . . . . . . . . . . . . .

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11 Boundary Control Systems 11.1 Boundary control systems . . . . . . . 11.2 Outputs for boundary control systems 11.3 Port-Hamiltonian systems as boundary 11.4 Exercises . . . . . . . . . . . . . . . . 11.5 Notes and references . . . . . . . . . .

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Contents

ix

12 Transfer Functions 12.1 Basic definition and properties . . . . . . . . . . 12.2 Transfer functions for port-Hamiltonian systems 12.3 Exercises . . . . . . . . . . . . . . . . . . . . . . 12.4 Notes and references . . . . . . . . . . . . . . . .

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157 158 163 167 169

13 Well-posedness 13.1 Well-posedness for boundary control systems 13.2 Well-posedness for port-Hamiltonian systems 13.3 P1 H diagonal . . . . . . . . . . . . . . . . . . 13.4 Proof of Theorem 13.2.2 . . . . . . . . . . . . 13.5 Well-posedness of the vibrating string . . . . 13.6 Exercises . . . . . . . . . . . . . . . . . . . . 13.7 Notes and references . . . . . . . . . . . . . .

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171 171 181 186 189 191 193 195

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A Integration and Hardy Spaces 197 A.1 Integration theory . . . . . . . . . . . . . . . . . . . . . . . . . . . 197 A.2 The Hardy spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . 202 Bibliography

209

Index

215

List of Figures 1.1 1.2 1.3 1.4 1.5 1.6 1.7

1.9

A system . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Electrical network . . . . . . . . . . . . . . . . . . . . . . . . . . . Mass-spring-system . . . . . . . . . . . . . . . . . . . . . . . . . . . The vibrating string . . . . . . . . . . . . . . . . . . . . . . . . . . Our system . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Feedback system . . . . . . . . . . . . . . . . . . . . . . . . . . . . The solution of the open loop system (1.30) and (1.29) (dashed line) and the solution of the closed loop system (1.30) and (1.27) (solid line) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . The solution of the system 2q (2) (t) = u(t) with feedback (1.27) (dashed line) and the solution of the system (1.30) with the same feedback (1.27) (solid line) . . . . . . . . . . . . . . . . . . . . . . . RCL network . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

9 10

3.1 3.2 3.3

Cart with pendulum . . . . . . . . . . . . . . . . . . . . . . . . . . Electrical network . . . . . . . . . . . . . . . . . . . . . . . . . . . Cart with two pendulums . . . . . . . . . . . . . . . . . . . . . . .

31 32 37

7.1 7.2 7.3

The vibrating string . . . . . . . . . . . . . . . . . . . . . . . . . . Transmission line . . . . . . . . . . . . . . . . . . . . . . . . . . . . Coupled vibrating strings . . . . . . . . . . . . . . . . . . . . . . .

79 94 95

8.1

Spectral decomposition . . . . . . . . . . . . . . . . . . . . . . . . . 102

9.1 9.2 9.3

The vibrating string with a damper . . . . . . . . . . . . . . . . . . 118 Transmission line . . . . . . . . . . . . . . . . . . . . . . . . . . . . 120 Coupled vibrating strings with dampers . . . . . . . . . . . . . . . 121

1.8

1 2 4 4 7 7

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11.1 The vibrating string with two controls . . . . . . . . . . . . . . . . 153 11.2 Coupled vibrating strings with external force . . . . . . . . . . . . 155 12.1 The vibrating string with two controls . . . . . . . . . . . . . . . . 166 12.2 Series connection . . . . . . . . . . . . . . . . . . . . . . . . . . . . 168 12.3 Parallel connection . . . . . . . . . . . . . . . . . . . . . . . . . . . 169 xi

xii

List of Figures 12.4 Feedback connection . . . . . . . . . . . . . . . . . . . . . . . . . . 169 13.1 The closed loop system . . . . . . . . . . . . . . . . . . . . . . . . . 180 13.2 The system (13.68) with input (13.71) and output (13.72) . . . . . 188

Chapter 1

Introduction In this chapter we provide an introduction to the field of mathematical systems theory. Besides examples we discuss the notion of feedback and we answer the question why feedback is useful. However, before we start with the examples we discuss the following picture, which can be seen as the essence of systems theory. In u

y P Figure 1.1: A system

systems theory, we consider models which are in contact with their environment. In the above picture, P denotes the to-be-studied-model which interacts with its environment via u and y. u and y are time signals, i.e., functions of the time t. The function u denotes the signal which influences P and y is the signal which we observe from P . u is called the input or control and y is called the output. The character P is chosen, since it is short for plant, think of chemical or power plant. To obtain a better understanding for the general setting as depicted in Figure 1.1, we discuss several examples in which we indicate the input and output. Regarding these examples, we should mention that we use different notations (1) for derivatives. In the best tradition of mathematics, we use f˙, df to dt , and f denote the first derivative of the function f . Similarly for higher derivatives.

1.1 Examples Example 1.1.1. Newton’s second law states that the force applied to a body produces a proportional acceleration; that is F (t) = m¨ q (t). B. Jacob and H.J. Zwart, Linear Port-Hamiltonian Systems on Infinite-dimensional Spaces, Operator Theory: Advances and Applications 223, DOI 10.1007/978-3-0348-0399-1_1, © Springer Basel 2012

(1.1) 1

2

Chapter 1. Introduction

Here q(t) denotes the position at time t of the particle with mass m, and F (t) is the force applied to it. Regarding the external force F (t) as our input u(t) and choosing the position q(t) as our output y(t), we obtain the differential equation y¨(t) =

1 u(t), m

t ≥ 0.

(1.2)

Thus the differential equation describes the behaviour “inside the box P ”, see Figure 1.1. Further, we can influence the system via the external force u and we observe the position y. In this simple example we clearly see that u is not the only quantity that determines y. The output also depends on the initial position q(0) and the initial velocity q(0). ˙ They are normally not at our disposal to choose freely, and so they are also “inside the box”. Example 1.1.2. Consider the electrical network given by Figure 1.2. Here V denotes the voltage source, L1 , L2 denote the inductance of the inductors, and C denotes the capacitance of the capacitor.

V L1

C

L2

Figure 1.2: Electrical network For the components in an electrical circuit, the following basic laws hold: the current IL and voltage VL across an inductor with inductance L is related via VL (t) = L

dIL (t), dt

(1.3)

whereas the current IC and voltage VC of the capacitor with capacitance C are related via dVC IC (t) = C (t). (1.4) dt The conservation of charge and energy in electrical circuits are described by Kirchhoff’s circuit laws. Kirchhoff’s first law states that at any node in an electrical circuit, the sum of currents flowing into the node is equal to the sum of currents flowing out of the node. Moreover, Kirchhoff’s second law says that the directed sum of the electrical potential differences around any closed circuit must be zero.

1.1. Examples

3

Applying these laws to our example, we obtain the following differential equations: dIL1 (t) = VL1 (t) = VC (t) + V (t), dt dIL2 L2 (t) = VL2 (t) = VC (t), and dt dVC C (t) = IC (t) = −IL1 (t) − IL2 (t). dt

(1.5)

L1

(1.6) (1.7)

We assume that we can only measure the current IL1 , i.e. we define y(t) = IL1 (t). Using (1.5), we obtain L1 y (1) (t) = VC (t) + V (t). Further, (1.7) then implies L1 Cy (2) (t) = −y(t) − IL2 (t) + CV (1) (t).

(1.8)

Differentiating this equation once more and using (1.6), we find dIL2 1 (t) + CV (2) (t) = −y (1) (t) − VC (t) + CV (2) (t) dt L2 1 = −y (1) (t) − L1 y (1) (t) − V (t) + CV (2) (t), (1.9) L2

L1 Cy (3) (t) = −y (1) (t) −

where we have used (1.5) as well. We regard the voltage supplied by the voltage source as the input u. Thus we obtain the following ordinary differential equation describing our system: L1 1 (3) L1 Cy (t) + 1 + y (1) (t) = u(t) + Cu(2) (t). (1.10) L2 L2 Example 1.1.3. Suppose we have a mass m which can move along a line, as depicted in Figure 1.3. The mass is connected to a spring with spring constant k, which in turn is connected to a wall. Furthermore, the mass is connected to a damper whose (friction) force is proportional to the velocity of the mass by the constant r. The third force which is working on the mass is given by the external force F (t). Let q(t) be the distance of the mass to the equilibrium point. Then by Newton’s law we have that m¨ q (t) = total sum of the forces. As the force of the spring equals kq(t) and the force by the damper equals rq(t), ˙ we find m¨ q (t) + rq(t) ˙ + kq(t) = F (t). (1.11) We regard the external force F (t) as our input u and we choose the position as our output y. This choice leads to the differential equation m¨ y (t) + ry(t) ˙ + ky(t) = u(t).

(1.12)

4

Chapter 1. Introduction q(t) k

m

F (t)

r Figure 1.3: Mass-spring-system Until now we have only seen examples which can be modelled by ordinary differential equations. The following examples are modelled via a partial differential equation.

11 00 00 11 00 11 00 11

Example 1.1.4. We consider the vibrating string as depicted in Figure 1.4. The u

Figure 1.4: The vibrating string string is fixed at the left-hand side and may move freely at the right-hand side. We allow that a force u may be applied at that side. The model of the (undamped) vibrating string is given by ∂2w 1 ∂ ∂w T (ζ) (ζ, t) (1.13) (ζ, t) = ∂t2 ρ(ζ) ∂ζ ∂ζ where ζ ∈ [a, b] is the spatial variable, w(ζ, t) is the vertical displacement of the string at position ζ and time t, T is the Young’s modulus of the string, and ρ is the mass density, which may vary along the string. This model is a simplified version of other systems where vibrations occur, as in the case of large structures, and it is also used in acoustics. The partial differential equation (1.13) is also known as the wave equation. If the mass density and the Young’s modulus are constant, then we get the partial differential equation ∂ 2w ∂ 2w (ζ, t) = c2 2 (ζ, t), 2 ∂t ∂ζ

(1.14)

1.1. Examples

5

where c2 = T /ρ. This is the most familiar form of the wave equation. In contrast to ordinary differential equations, we need boundary conditions for our partial differential equation (1.13) or (1.14). At the left-hand side, we put the position to zero, i.e., w(a, t) = 0 (1.15) and at the right-hand side, we have the balance of the forces, which gives T (b)

∂w (b, t) = u(t). ∂ζ

(1.16)

There are different options for the output. One option is to measure the velocity at the right-hand side, i.e., ∂w y(t) = (b, t). (1.17) ∂t Another option could be to measure the velocity at a point between a and b, or to measure the position of the wave, i.e., y(t) = w(·, t). Hence at every time instant, y is a function of the spatial coordinate. We end this section with another well-known partial differential equation. Example 1.1.5. The model of heat conduction consists of only one conservation law, that is, the conservation of energy. It is given as ∂e ∂ = − JQ , ∂t ∂ζ

(1.18)

where e(ζ, t) is the energy density and JQ (ζ, t) is the heat flux. This conservation law is completed by two closure equations. The first one expresses the calorimetric properties of the material: ∂e = cV (T ), (1.19) ∂T where T (ζ, t) is the temperature distribution and cV is the heat capacity. The second closure equation defines the heat conduction property of the material (Fourier’s conduction law): JQ = −λ(T, ζ)

∂T , ∂ζ

(1.20)

where λ(T, ζ) denotes the heat conduction coefficient. Assuming that the variations of the temperature are not too large, we may assume that the heat capacity and the heat conduction coefficient are independent of the temperature. Thus we obtain the partial differential equation ∂T 1 ∂ ∂T (ζ, t) = (ζ, t) , ζ ∈ (a, b), t ≥ 0. (1.21) λ(ζ) ∂t cV ∂ζ ∂ζ

6


As for the vibrating string, the constant coefficient case is better known. This is ∂T ∂2T (ζ, t) = α 2 (ζ, t), ∂t ∂ζ

ζ ∈ (a, b), t ≥ 0

(1.22)

with α = λ/cV . Again, we need boundary conditions for the partial differential equations (1.21) and (1.22). If the heat conduction takes places in a perfectly insulated surrounding, then no heat can flow in or out of the system, and we have as boundary conditions ∂T ∂T (a, t) = 0, and λ(b) (b, t) = 0. (1.23) λ(a) ∂ζ ∂ζ It can also be that the temperature at the boundary is prescribed. For instance, if the ends are lying in a bath with melting ice, then we obtain the boundary conditions T (a, t) = 0, and T (b, t) = 0. (1.24) As measurement we can take the temperature at a point y(t) = T (ζ0 , t), ζ0 ∈ (a, b). Another choice could be the average temperature in an interval (ζ0 , ζ1 ). In the latter case we find ζ1 1 T (ζ, t) dζ. y(t) = ζ1 − ζ0 ζ0 As input we could control the temperature at one end of the spatial interval, e.g. T (b, t) = u(t), or we could heat it in the interval (ζ0 , ζ1 ). The latter choice leads to the partial differential equation 1 ∂ ∂T ∂T (ζ, t) = (ζ, t) + u(ζ, t), λ(ζ) ∂t cV ∂ζ ∂ζ where we define u(ζ, t) = 0 for ζ ∈ (ζ0 , ζ1 ).

1.2 How to control a system? In the previous section we have seen that there are many models in which we can distinguish an input and an output. Via the input we have the possibility to influence the system. In particular, we aim to choose the input u such that y or all variables in the box behave as we desire. Note that the phrase “to behave as we desire” means that we have to make choices. These choices will be based on the type of plant we are dealing with. If P represents a passenger airplane, and y the height, we do not want y to go from 10 kilometers to ground level in one second. However, we would like that this happens in half an hour. On the other hand, if the plant represents a stepper, then very fast action is essential. A stepper is a device used in the manufacture of integrated circuits (ICs); it got its name

1.2. How to control a system?

7

from the fact that it moves or “steps” the silicon wafer from one shot location to another. This has to be done very quickly, and with a precision on nano-meters. So generally, the control task is to find an input u such that y has a desired behaviour and we shall rarely do it by explicitly calculating the function u. More often we design u on the basis of y. Hence instead of open loop systems described by Figure 1.1, here given once more as Figure 1.5, we work with closed loop systems given by Figure 1.6. In order to “read” the latter picture, it is sufficient to know y

u

P

Figure 1.5: Our system

u

y P

C

Figure 1.6: Feedback system some simple rules. As before, by a rectangular block we denote a model relating the incoming signal to the outgoing signal. This could for instance be an ordinary or partial differential equation. Mathematically speaking one may see P and C as operators mapping the signal u to the signal y and vice versa. At a node we assume that the incoming signals are the same as the outgoing signals. The arrows indicate the directions of the signals. As before we denote by P the system that we desire to control, and by C we denote our (designed) controller. In this section we show that closed loop systems have in general better properties than open loop systems. We explain the advantages by means of an example. We consider the simple control problem of steering the position of the mass m to zero, see Example 1.1.1. Hence, the control problem is to design the force F such that, for every initial position and every initial velocity of the mass, the position of the mass is going to zero for time going to infinity. To simplify the problem even more we assume that the mass m equals 1, and we assume that we measure the velocity and the position, i.e., q(t) q¨(t) = u(t) and y(t) = . (1.25) q(t) ˙

8


In Exercise 1.2 the case y(t) = q(t) is discussed. In order to design the controller C for the plant described by (1.25) we proceed as follows. First we have to specify what we mean by “to behave as we desire”, that is, we have to decide how and how fast the position should go to zero. It is quite common to choose an “ideal” differential equation such that the solutions have the desired behaviour. We consider the solutions of the differential equation f (2) (t) + 2f (1) (t) + f (t) = 0. (1.26) The general solution of (1.26) is given by f (t) = αe−t + βte−t . This is to our satisfaction, and so we try to design a control u such that the position q is equal to a solution of the equation (1.26). If we choose

u(t) = −1 −2 y(t) = −2q (1) (t) − q(t), (1.27) then the position of the mass indeed behaves in the same way as the solutions of (1.26). Note that for the design of this feedback law no knowledge of the initial position nor the initial velocity is needed. If we want to obtain the same behavior of the solution by an open loop control, i.e., if we want to construct the input as an explicit function of time, we need to know the initial position and the initial velocity. Suppose they are given as q(0) = −2 and q (1) (0) = 5, respectively. Calculating the position as the solution of q (2) (t) + 2q (1) (t) + q(t) = 0, we obtain Thus by (1.27) we find

t ≥ 0,

q(0) = −2,

q (1) (0) = 5,

q(t) = −2e−t + 3te−t .

(1.28)

u(t) = −8e−t + 3te−t .

(1.29)

Applying this input to the system q (2) (t) = u(t)

(1.30)

would give the same behavior as applying (1.27). We simulate the open loop system, i.e., (1.30) with u(t) given by (1.29) and the closed loop system, i.e, (1.30) with u(t) given by (1.27). The result is shown in Figure 1.7. We obtain that the simulation of the open loop system is worse than the one of the closed loop system. This could be blamed on a bad numerical solver, but even mathematically, we can show that the closed loop system behaves in a superior manner to the open loop system. We have assumed that we know the initial data exactly, but this will never be the case. So suppose that we have (small) errors in both initial conditions, but we are unaware of the precise error. Thus we apply the input (1.29) to the system q (2) (t) = u(t),

t ≥ 0,

q(0) = q0 ,

q (1) (0) = q1 ,

1.2. How to control a system?

9

1

0.5

y→

0

-0.5

-1

-1.5

-2 0

2

4

6

8

10

t→

12

14

16

18

20

Figure 1.7: The solution of the open loop system (1.30) and (1.29) (dashed line) and the solution of the closed loop system (1.30) and (1.27) (solid line) 2.5 2 1.5

q→

1 0.5 0

-0.5 -1 -1.5 -2 0

2

4

6

8

10

t→

12

14

16

18

20

Figure 1.8: The solution of the system 2q (2) (t) = u(t) with feedback (1.27) (dashed line) and the solution of the system (1.30) with the same feedback (1.27) (solid line) where q0 and q1 are not exactly known initial conditions. The solution of this ordinary differential equation is given by q(t) = −2e−t + 3te−t + (q0 + 2) + (q1 − 5)t.

(1.31)

Hence any small error in q(0) will remain, and any error in q (1) (0) will even increase. This effect also occurred in the simulation. No matter how good the quality

10


of the simulation is, there will always be small errors, which results in the misfit of the solution. Apart from the initial conditions which could contain a (small) error, there is always the question of the exact value of the physical parameters. To illustrate this, assume that we have measured the mass with a large error, so assume that the real value of the mass is 2. By means of a simulation, we show that the feedback law (1.27) still works. Note, the function q can also be calculated analytically. The position is still converging to zero, and so the design criterion is still satisfied. This shows the power of feedback.

1.3 Exercises 1.1. Consider the electrical network as given below.

V

L

R

C

Figure 1.9: RCL network As input we choose the voltage supplied by the voltage source, and as output we measure the voltage over the resistor. Note that the current IR and the voltage VR across a resistor with resistance R is related via VR = RIR . Determine the ordinary differential equation modeling the relation between u and y. 1.2. As in Section 1.2 we study the system q¨(t) = u(t).

(1.32)

In this and the next exercise we investigate the properties of the feedback a little further. First we show that the feedback as designed in Section 1.2 still works well if we add some perturbation to the output. In the second part we study whether the same holds for the open loop system. (a) As output we choose

y(t) =

q(t) q(t) ˙

.

However, we assume that we cannot observe the output exactly due to some noise. We model the noise as a small signal which changes quickly,

1.3. Exercises

11

and so we take εε sin(ωt) sin(ωt) , with ε small, and ω large. For simplicity we have taken both components of the noise to be equal, but this is non-essential. So the output that we have at our disposal for controlling the system is q(t) + εε sin(ωt) . q(t) ˙ sin(ωt) We apply the same feedback law as given in (1.27), that is, u(t) = −2(q (1) (t) + ε sin(ωt)) − (q(t) + ε sin(ωt)). Determine the solution of the closed loop system, and conclude that the perturbation on the output y remains bounded. More precisely, the perturbation on the output y is bounded by C ωε , where C > 0 is independent of ε > 0 and ω ≥ 1. (b) In the previous part we have seen that adding noise to the observation or control hardly effects the desired behavior when we work with feedback systems. We now investigate if the same holds for the open loop system. We consider the system (1.32) with initial conditions q(0) = −2, q(0) ˙ = 5, as input we take u(t) = −8e−t + 3te−t − 3ε sin(ωt) and as output we choose

y(t) =

q(t) q(t) ˙

.

Calculate the output y and conclude that the perturbation on the output y is unbounded. 1.3. We consider again the system described by equation (1.32), but now we assume that we measure the position only, i.e., y(t) = q(t). (a) As feedback we choose u(t) = ky(t), k ∈ R, where y(t) = q(t). Show that for every k ∈ R there exists an initial condition such that the solutions of the differential equation (1.32) with this feedback do not converge to zero. (b) As in the previous part we choose the output y(t) = q(t), but we model the controller C again by a differential equation, that is, the input u is an output of another system. More precisely, we choose z(t) ˙ = −3z(t) + 8y(t),

u(t) = z(t) − 3y(t).

Show that if y, z, u satisfies z(t) ˙ = −3z(t) + 8y(t), u(t) = z(t) − 3y(t), y¨(t) = u(t),

(1.33)

12

Chapter 1. Introduction then y satisfies the equation y (3) (t) = z (1) (t) − 3y (1) (t) = −3 y (2) (t) + 3y(t) + 8y(t) − 3y (1) (t). Conclude that y converges exponentially to zero. More precisely, we have |y(t)| ≤ C(t2 + 1)e−t , for some constant C > 0.

1.4 Notes and references The examples and results presented in this chapter can be found in many books on systems theory. We refer to [50] and [32] for the examples described by ordinary differential equations, and to [10] for the examples described by partial differential equations. [50] also provides a discussion on feedback systems.

Chapter 2

State Space Representation In the previous chapter we introduced models with an input and an output. These models were described by an ordinary or partial differential equation. However, there are other possibilities to model systems with inputs and outputs. In this chapter we introduce the state space representation on a finite-dimensional state space. Later we will encounter these representations on an infinite-dimensional state space. State space representations enable us to study systems with inputs and outputs in a uniform framework. In this chapter, we show that every model described by an ordinary differential equation possesses a state space representation on a finite-dimensional state space, and that it is just a different way of writing down the system. However, this different representation turns out to be very important as we will see in the following chapters. In particular, it enables us to develop general control strategies.

2.1 State space models In this section, we show that every linear ordinary differential equation with constant coefficients can be written in the form x(t) ˙ = Ax(t) + Bu(t),

y(t) = Cx(t) + Du(t),

(2.1)

where x is a vector in Rn (or Cn ), and A, B, C, and D are matrices of appropriate sizes. This is known as the state space representation, state space system or state space model. The vector x is called the state. State space representations with a finite-dimensional state space are also called finite-dimensional systems. The first equation of (2.1) is named the state differential equation. Note that most of the calculations in this section are formally, that is, we do not need the notion of (classical) solution of linear ordinary differential equations with constant coefficients nor of a state space representation. Further, we have no specific assumption on the input u. One may for simplicity just assume that B. Jacob and H.J. Zwart, Linear Port-Hamiltonian Systems on Infinite-dimensional Spaces, Operator Theory: Advances and Applications 223, DOI 10.1007/978-3-0348-0399-1_2, © Springer Basel 2012

13

14

Chapter 2. State Space Representation

u is sufficiently smooth. The aim of this section is to show that under some mild conditions state space representations and linear ordinary differential equations with constant coefficients are equivalent. Note that, under mild assumptions, it can be shown that both representations possess the same solutions. In the next section, we discuss the notion of classical and mild solutions of the state space representation (2.1) in more details. First of all we show that the Examples 1.1.1 and 1.1.2 have a state space representation. Example 2.1.1. Consider Newton’s law of Example 1.1.1, with input the force and output the position, see equation (1.2), u(t) = m¨ y (t). We choose the state x(t) ∈ R2 as x(t) =

y(t) y(t) ˙

.

For this choice we see that 0 y(t) ˙ 0 1 y(t) x(t) ˙ = = + u(t). 1 y¨(t) 0 0 y(t) ˙ m

(2.2)

Thus Newton’s law can be written in the standard state space formulation with

0 1 0 , C= 1 0 and D = 0. A= , B= 1 0 0 m For Newton’s law we had to introduce a new state variable in order to obtain the state space representation. However, some models appear naturally in the state space representation. This holds for example for the electrical circuit of Example 1.1.2. Example 2.1.2. Consider the electrical circuit of Example 1.1.2. The differential equation relating the voltage V provided by the voltage source, and the current through the first inductor IL1 is given by the differential equation L1 1 (3) L1 Cy (t) + 1 + y (1) (t) = u(t) + Cu(2) (t). L2 L2 However, to find the state space representation we use the equations (1.5)–(1.7) directly, that is, we consider the equations dIL1 (t) = VL1 (t) = VC (t) + V (t), dt dIL2 (t) = VL2 (t) = VC (t), and L2 dt dVC (t) = IC (t) = −IL1 (t) − IL2 (t). C dt L1

(2.3) (2.4) (2.5)

2.1. State space models As state vector we choose

15

⎡

⎤ ⎡ ⎤ x1 (t) IL1 (t) x(t) = ⎣ x2 (t) ⎦ = ⎣ IL2 (t) ⎦ . x3 (t) VC (t)

Using the equations (2.3)–(2.5) we find ⎤ ⎡ ⎡ 1 1 0 L1 x3 (t) + L1 V (t) 1 ⎦ ⎣ ⎣ 0 x(t) ˙ = = L2 x3 (t) − C1 x1 (t) − C1 x2 (t) − C1

⎤⎡

⎤ ⎡ 1 ⎤ x1 (t) L1 ⎦ ⎣ x2 (t) ⎦ + ⎣ 0 ⎦ V (t). x3 (t) 0 0 (2.6) If we use the (standard) notation u for the input V and y for the output/measurement IL1 , then the system of Example 1.1.2 can be written in the state space representation with ⎡ ⎡ 1 ⎤ 1 ⎤ 0 0 L1 L1

1 ⎦ 0 , B = ⎣ 0 ⎦ , C = 1 0 0 and D = 0. A=⎣ 0 L2 0 − C1 − C1 0 0 0 − C1

1 L1 1 L2

Our construction of the state space representation may seem rather ad-hoc. However, next we show that every linear ordinary differential equation with constant coefficients has a state space representation. In order to explain the main ideas we first consider an example. Example 2.1.3. Consider the following differential equation. y¨(t) + 5y(t) ˙ + 6y(t) = 7u(t) ˙ + 8u(t).

(2.7)

The standard way of deriving a state space representation from a differential equation is to transform it into an integral equation. In order to this we first move every term, except the highest derivative of y, to the right-hand side of the equation. Thus we obtain, y¨(t) = 7u(t) ˙ + 8u(t) − 5y(t) ˙ − 6y(t). Now we integrate this equation as often as needed to remove all the derivatives of y, here we have to integrate twice t s 7u(τ ˙ ) + 8u(τ ) − 5y(τ ˙ ) − 6y(τ ) dτ ds + y(0) + y(0)t ˙ y(t) = 0 0 t s = 7u(s) − 5y(s) − 7u(0) + 5y(0) + 8u(τ ) − 6y(τ ) dτ ds + y(0) + y(0)t ˙ 0 0 t s = y(0) + 7u(s) − 5y(s) − 7u(0) + 5y(0) + y(0) ˙ + 8u(τ ) − 6y(τ ) dτ ds. 0

0

(2.8)

16


Note that we only have to evaluate two integrals: the integral of −6y(t) + 8u(t), and the integral of the first integral plus 7u(t) − 5y(t) − 7u(0) + 5y(0) + y(0). ˙ Up to a constant, we choose the result of these integrals as our state variables, i.e., xk (t), k = 1, 2, t s y(t) = 7u(s) − 5y(s) −7u(0) + 5y(0) + y(0) ˙ + 8u(τ ) − 6y(τ ) dτ ds + y(0) . 0 0

x2 (s)

x1 (t)

With this choice of variables we obtain ⎧ ⎨ y(t) = x1 (t) x˙ 1 (t) = 7u(t) − 5y(t) + x2 (t) , ⎩ x˙ 2 (t) = 8u(t) − 6y(t) or equivalently

−5 1 7 x(t) ˙ = x(t) + u(t), −6 0 8

y(t) = 1 0 x(t).

(2.9) (2.10)

This is a state space representation of the differential equation (2.7). For a general differential equation with constant coefficients we can adopt the procedure used in the previous example. In order not to drown in the notation and formulas we omit the variable t. Consider the differential equation y (n) + pn−1 y (n−1) + · · · + p1 y (1) + p0 y = qm u(m) + qm−1 u(m−1) + · · · + q1 u(1) + q0 u. (2.11) We assume that n ≥ m. Note that without loss of generality, we may assume that m = n, otherwise we may choose qm+1 = · · · = qn = 0. Moving every y-term, except the highest derivative of y, to the right-hand side gives y (n) = qn u(n) + qn−1 u(n−1) − pn−1 y (n−1) +· · ·+ q1 u(1) − p1 y (1) +(q0 u − p0 y) . In order to explain the main ideas we ignore the initial conditions in the following. Integrating this equation n times and reordering the terms, we obtain y = qn u + qn−1 u − pn−1 y + · · · + q1 u − p1 y + (q0 u − p0 y) . Each integral defines a state variable y = qn u + qn−1 u − pn−1 y + · · · + q1 u − p1 y + (q0 u − p0 y) .

x1

xn−1

xn

2.1. State space models

17

Hence we obtain the following (differential) equations: y(t) = qn u(t) + x1 (t), x˙ 1 (t) = qn−1 u(t) − pn−1 y(t) + x2 (t), .. .. . = . x˙ n−1 (t) = q1 u(t) − p1 y(t) + xn (t), x˙ n (t) = q0 u(t) − p0 y(t). This system corresponds to the state space representation ⎤ ⎡ ⎡ −pn−1 1 0 · · · 0 qn−1 − pn−1 qn ⎢ ⎥ ⎢ . .. .. . . . . .. ⎥ .. ⎢ ⎢ . . ⎥ ⎢ ⎢ ⎥ x(t) + ⎢ . .. x(t) ˙ =⎢ . ⎢ ⎢ . 1 0 ⎥ . ⎢ ⎥ ⎢ ⎣ −p1 ⎣ 0 ··· 0 1 ⎦ q1 − p1 qn −p0 0 ··· 0 0 q0 − p0 qn

y(t) = 1 0 . . . 0 x(t) + qn u(t).

⎤ ⎥ ⎥ ⎥ ⎥ u(t), ⎥ ⎥ ⎦

(2.12)

(2.13)

We summarize the above result in a theorem. Theorem 2.1.4. If m ≤ n, then the ordinary differential equation (2.11) can be written as a state space system with an n-dimensional state space. One possible choice is given by (2.12) and (2.13). Concluding, every ordinary differential equation with constant coefficients has a space space representation, provided the highest derivative of the input does not exceed the highest derivative of the output. A natural question is, if the converse holds as well. Hence given a state space representation, can we find an ordinary differential equation relating u and y. We show next that this holds under a mild condition. Theorem 2.1.5. Consider the state space system as given in equation (2.1) and assume that the output is scalar-valued. If the matrix ⎡ ⎤ C ⎢ CA ⎥ ⎥ ⎢ (2.14) O=⎢ ⎥ .. ⎦ ⎣ . CAn−1 has full rank, then there exists a differential equation describing the relation of u and y. Proof. In order to derive the differential equation describing the relation of u and y, we assume that the input u is (n − 1)-times differentiable. This implies that the output y is also (n − 1)-times differentiable. As the output is scalar-valued, the

18


matrix O is a square matrix. Thus by assumption O is invertible. Differentiating the equation y(t) = Cx(t) + Du(t) and using the differential equation for x, we find y (1) = C (Ax + Bu) + Du(1) . By induction it is now easy to see that y (k) = CAk x +

k−1

CA Bu(k−1−) + Du(k) .

(2.15)

=0

Hence we have that ⎡ ⎤ ⎡ y ⎢ y (1) ⎥ ⎢ ⎢ ⎥ ⎢ ⎢ ⎥=⎢ .. ⎣ ⎦ ⎣ .

⎤

C CA .. . CAn−1 ⎡

y (n−1)

⎢ ⎢ = Ox + ⎢ ⎢ ⎣

⎡

D

0

CB .. .

D .. .

CAn−2 B

···

⎢ ⎥ ⎢ ⎥ ⎥x + ⎢ ⎢ ⎦ ⎣ D

0

CB .. .

D .. . ···

CAn−2 B

··· .. . .. . CB

Since O is invertible, x can be expressed by u, y ⎡ ⎡ ⎤ D 0 y ⎢ ⎢ y (1) ⎥ ⎢ CB D ⎢ ⎥ x = O−1 ⎢ ⎥ − O−1 ⎢ .. ⎢ . . ⎣ ⎦ .. .. . ⎣ n−2 y (n−1) B ··· CA

0 .. . 0 D

⎤ ··· 0 ⎡ u .. ⎥ ⎢ u(1) .. ⎥ . . ⎥⎢ .. ⎥⎢ .. . . 0 ⎦⎣ u(n−1) CB D ⎤⎡ ⎤ u ⎥⎢ ⎥ ⎢ u(1) ⎥ ⎥ ⎥⎢ ⎥. .. ⎥⎣ ⎦ . ⎦ (n−1) u

and its derivatives ⎤⎡ ··· 0 u .. ⎥ ⎢ u(1) .. . . ⎥ ⎥⎢ .. ⎥⎢ .. . . 0 ⎦⎣ (n−1) u CB D

⎤ ⎥ ⎥ ⎥ ⎦

(2.16)

⎤ ⎥ ⎥ ⎥ . (2.17) ⎦

Inserting this in (2.15) with k = n, we find y (n) = CAn x +

n−1

CA Bu(n−1−) + Du(n)

=0

⎡

⎢ ⎢ = CAn O−1 ⎢ ⎣

y y (1) .. . y (n−1)

+

n−1

⎤

⎡

⎢ ⎥ ⎢ ⎥ ⎥ − CAn O−1 ⎢ ⎢ ⎦ ⎣

CA Bu(n−1−) + Du(n) .

D

0

CB .. .

D .. .

CAn−2 B

···

··· .. . .. . CB

⎤⎡ 0 u .. ⎥ ⎢ u(1) ⎥ . ⎥⎢ .. ⎥⎢ ⎣ . 0 ⎦ u(n−1) D

⎤ ⎥ ⎥ ⎥ ⎦

(2.18)

=0

This is a differential equation relating u and y, and thus the theorem is proved.

2.2. Solutions of the state space models

19

We apply the above result on the example of the electrical circuit, see Example 2.1.2. Example 2.1.6. Consider the state space representation of Example 2.1.2. Since n = 3, the matrix O, see (2.14) is given by ⎡ ⎤ 1 0 0 1 ⎦ 0 0 O=⎣ L1 1 1 − L1 C − L1 C 0 which is invertible with inverse O−1

⎡

1 0 = ⎣ −1 0 0 L1

⎤ 0 −L1 C ⎦ . 0

We calculate the other terms in equation (2.18) 1 1 1 1 , CAB = 0, and CA2 B = − 2 . CA3 = 0 0 − L21 C − L1 L2 C , CB = L1 L1 C Substituting this in (2.18), we obtain y

(3)

=

− +

0

− L21C 1

0

0

0

−

− L21C 1

1 L1 L2 C

−

1 L1 L2 C

1 1 (2) u − 2 u. L1 L1 C

⎡

⎤⎡ ⎤ y 1 0 0 (1) ⎣ −1 0 −L1 C ⎦ ⎣ y ⎦ 0 L1 0 y (2) ⎡ ⎤⎡ 0 0 1 0 0 ⎣ −1 0 −L1 C ⎦ ⎣ L1 0 1 1 0 0 L1 0 L1

⎤⎡ ⎤ 0 u (1) ⎦ 0 ⎦⎣ u 0 u(2)

Evaluating all the products, we find the differential equation 1 1 1 1 1 (2) 1 (3) (1) y =− + + y − u+ u − 2 u L1 C L2 C L21 C L1 L2 C L1 L1 C 1 1 (2) 1 1 =− + y (1) + u + u. (2.19) L1 C L2 C L1 L1 L2 C This is exactly equation (1.10). Hence, we have shown that state space models appear naturally when modeling a system, but they can also be obtained from an ordinary differential equation.

2.2 Solutions of the state space models In this section, we derive the solution of equation (2.1). This equation consists of a differential equation and an algebraic equation. If the differential equation

20


is solved, then the solution of the algebraic equation follows directly. Thus we concentrate here on the differential equation x(t) ˙ = Ax(t) + Bu(t),

x(0) = x0 .

(2.20)

In our examples all our matrices and state spaces were real. However, sometimes it is useful to work on Cn instead of Rn . So we assume that A ∈ Kn×n , B ∈ Kn×m and x0 ∈ Kn , where K equals either R or C. Equation (2.20) is a system of linear inhomogeneous differential equations with constant coefficients. By a classical solution of (2.20) we mean a function x ∈ C 1 ([0, ∞); Kn ) satisfying (2.20). Theorem 2.2.1. If u ∈ C([0, ∞); Km ), then the unique classical solution of (2.20) is given by the variation of constant formula t x(t) = eAt x0 + eA(t−s) Bu(s) ds, t ≥ 0. (2.21) 0

For completeness we include a proof of this theorem. However, we assume that the reader is familiar with the fact that the (unique) solution of x(t) ˙ = Ax(t), x(0) = x0 is given by eAt x0 . Proof. We first assume that x is a classical solution of (2.20). Then we have d A(t−s) [e x(s)] = eA(t−s) x(s) ˙ − AeA(t−s) x(s) ds = eA(t−s) [Ax(s) + Bu(s)] − AeA(t−s) x(s) = eA(t−s) Bu(s), which implies t eA(t−s) Bu(s) ds = 0

t 0

d A(t−s) [e x(s)] ds = eA(t−t) x(t) − eA(t−0) x(0) ds

= x(t) − eAt x0 . Equivalently, equation (2.21) holds. This shows that the solution is unique provided it exists. Thus it remains to show the existence of a classical solution. Clearly, all we need to show is that x defined by (2.21) is an element of C 1 ([0, ∞); Kn ) and satisfies the differential equation (2.20). It is easy to see that x ∈ C 1 ([0, ∞); Kn ). Moreover, for t ≥ 0 we have t x(t) ˙ = AeAt x0 + Bu(t) + AeA(t−s) Bu(s) ds 0 t eA(t−s) Bu(s) ds + Bu(t) = A eAt x0 + 0

= Ax(t) + Bu(t). This concludes the proof of the theorem.

2.3. Port-Hamiltonian systems

21

Thus for continuous input signals, there exists always a unique classical solution of the differential equation (2.20). However, we would like to have the freedom to deal with discontinuous and in particular (square) integrable inputs. Hence we choose as our input function space the set of locally integrable functions, L1loc ([0, ∞); Km ). Under mild conditions the existence of a classical solution implies the continuity of the input function u, see Exercise 2.2. Thus for u ∈ L1loc ([0, ∞); Km ) we cannot expect that the differential equation (2.20) possesses a classical solution. However, for these input functions the integral (2.21) is well-defined, and it defines a mild solution as we show in the following. Definition 2.2.2. A continuous function x : [0, ∞) → Kn is called a mild solution of (2.20) if x is continuous and satisfies the integrated version of the differential equation (2.20), i.e., if it satisfies

t

x(t) = x0 +

Ax(s) + Bu(s) ds

for t ≥ 0.

(2.22)

0

The following theorem is proved in Exercise 2.2. Theorem 2.2.3. Let u ∈ L1loc ([0, ∞); Km ). Then equation (2.20) possesses a unique mild solution which is given by (2.21). Thus for every u ∈ L1loc ([0, ∞); Km ) the state space system (2.1) has the unique (mild) solution given by

t

x(t) = eAt x0 +

eA(t−s) Bu(s) ds, 0

y(t) = CeAt x0 +

t

CeA(t−s) Bu(s) ds + Du(t). 0

In the next session we discuss a special class of state space models.

2.3 Port-Hamiltonian systems The state space of the state space system (2.1) is given by the Euclidean space Rn or Cn , and thus it is natural to work with the standard Euclidean norm/inner product on the state space. However, for many (physical) examples this is not the best choice. For these systems it is preferable to take the energy as the norm. To illustrate this, we return to Example 2.1.2. Example 2.3.1. The model of the electrical network of Example 2.1.2 is given by, see (2.3)–(2.5),

22

Chapter 2. State Space Representation dIL1 (t) = VL1 (t) = VC (t) + V (t), dt dIL2 L2 (t) = VL2 (t) = VC (t), dt dVC (t) = IC (t) = −IL1 (t) − IL2 (t). C dt L1

(2.23) (2.24) (2.25)

As in Example 2.1.2 we choose IL1 , IL2 , and VC as the state space variables, and so the natural norm on the state space seems to be ⎡ ⎤ IL1 ⎣ IL2 ⎦ = I 2 + I 2 + V 2 . L1 L2 C VC However, this is not the best choice. The preferred norm for this example is the square root of the energy of the (physical) system. For this particular example, the energy equals 1 E= L1 IL2 1 + L2 IL2 2 + CVC2 . (2.26) 2 Although this norm is equivalent to the Euclidean norm, the energy norm has some advantages. To illustrate this, we differentiate the energy along solutions. Using (2.23)–(2.25), we find that dIL1 dIL2 dE dVC (t) = L1 (t)IL1 (t) + L2 (t)IL2 (t) + C (t)VC (t) dt dt dt dt = (VC (t) + V (t))IL1 (t) + VC (t)IL2 (t) + (−IL1 (t) − IL2 (t))VC (t) (2.27) = V (t)IL1 (t). Thus the derivative of the energy along solutions equals the product of the input times the output, see Example 2.1.2. Note that, if we apply no voltage to the system, then the energy will remain constant which implies that the eigenvalues of the matrix A in Example 2.1.2 are on the imaginary axis. The energy of the system does not only provide a link with physics, also system properties become simpler to prove. The above example is a particular example of a port-Hamiltonian system. Recall that a matrix H is positive-definite if it is self-adjoint and if x∗ Hx > 0 for all vectors x = 0. Note that we always write x∗ even if we deal with real vectors, and in this situation x∗ equals the transpose of the vector. Definition 2.3.2. Let H be a positive-definite matrix, and let J be a skew-adjoint matrix, i.e., J ∗ = −J. Then the system x(t) ˙ = JHx(t) + Bu(t) ∗

y(t) = B Hx(t)

(2.28) (2.29)


23

is called a port-Hamiltonian system associated to H and J. J is called the structure matrix and H is the Hamiltonian density. The Hamiltonian associated to H is 1 ∗ 2 x Hx. This definition is a special case of a much more general definition. We have restricted ourselves here to the linear case, in which the Hamiltonian is quadratic, i.e., equals 12 x∗ Hx, but other Hamiltonian’s are also possible. In many examples the Hamiltonian will be equal to the energy of the system. Since H is positive-definite, the expression 12 x∗ Hx defines a new norm on Kn , see for example (2.26). This norm is associated to the inner product x, y H := y ∗ Hx, and is equivalent to the Euclidean norm. Port-Hamiltonian systems possess the following properties. Lemma 2.3.3. Let the norm · H on Kn be defined as x H := 12 x∗ Hx, where H is an arbitrary positive-definite matrix, and let x be a (classical) solution of (2.28)–(2.29). Then the following equality holds: d x(t) 2H = Re (y(t)∗ u(t)) . (2.30) dt Proof. As x is a classical solution, we may differentiate the squared norm, and obtain d x(t) 2H 1 1 = x(t)∗ Hx(t) ˙ + x(t) ˙ ∗ Hx(t) dt 2 2 1 1 ∗ = x(t)∗ H (JHx(t) + Bu(t)) + (JHx(t) + Bu(t)) Hx(t) 2 2 1 = (x(t)∗ HJHx(t)+x(t)∗ H∗ J ∗ Hx(t)+x(t)∗ HBu(t)+u∗(t)B ∗ Hx(t)) 2 1 = 0 + (y(t)∗ u(t) + u∗ (t)y(t)) , (2.31) 2 where we have used that J is skew-adjoint and H is self-adjoint. From the equality (2.30), we can conclude several facts. If u ≡ 0, then the Hamiltonian is constant. Thus the solutions lie on isoclines, and the energy remains constant when applying no control. Furthermore, we also get an idea of how to stabilize the system, i.e., how to steer the state to zero. The input u(t) = −ky(t), k ≥ 0, makes the energy non-increasing. This line of research will be further developed in Chapter 4. We end this chapter by identifying the structure matrix and the Hamiltonian density for the electrical network of Example 2.3.1. If we choose the state x as in Example 2.1.2, then ⎤ ⎤ ⎡ ⎡ 1 0 0 L1 0 0 L1 C 1 ⎦ , H = ⎣ 0 L2 0 ⎦ . 0 0 J =⎣ (2.32) L2 C 1 1 0 0 C − L1 C − L2 C 0 However, it is possible to choose a state variable for which J is not depending on the physical parameters, see Exercise 2.3.

24


2.4 Exercises 2.1. In this exercise we show that a state space representation is not uniquely determined. (a) Consider Example 2.1.1, and obtain a state space representation of Newton’s law with state x(t) =

y(t) ˙ y(t)

.

(b) Consider the state space model (2.1). Let T by an invertible n × nmatrix, and define z(t) = T x(t). Show that z is also a state of a state space representation, i.e., determine AT , BT , CT , and DT such that z(t) ˙ = AT z(t) + BT u(t)

y(t) = CT z(t) + DT u(t).

(2.33)

(c) Under the assumption that the output function is scalar-valued and the matrix O, see (2.14), has full rank, show that the differential equation relating u and y determined by (2.1) equals the differential equation relating u and y determined by (2.33). Hence although the state is non-unique, the differential equation relating the input and output is. 2.2. In this exercise we study the solutions of the state space representation (2.20) in more detail. (a) Show that if x is a classical solution of (2.20) and B is injective, then u is continuous. (b) Prove Theorem 2.2.3. Beside the notion of a mild solution there is also the notion of a weak solution. A function x : [0, ∞) → Kn is said to be a weak solution of the differential equation (2.20), if x is continuous and if for every t > 0 and every g ∈ C 1 ([0, t]; Kn ) the following holds t g(τ )∗ (Ax(τ ) + Bu(τ )) + g(τ ˙ )∗ x(τ ) dτ = g(t)∗ x(t) − g(0)∗ x0 . 0

It can be proved that for every u ∈ L1loc ([0, ∞); Km ), the function x given by (2.21) is the unique weak solution of (2.20). 2.3. In this exercise we investigate again the port-Hamiltonian structure of the electrical network of Example 2.1.2. (a) Show that the state space model as derived in Example 2.1.2 can be written as a port-Hamiltonian model with the J and H given by (2.32). (b) For a port-Hamiltonian system the matrix A in the state space representation is given by JH. However, there are many ways of writing

2.5. Notes and references

25

a matrix as a product of two other matrices, and thus the matrices J and H are not uniquely determined. Therefore, we may add additional conditions. The standard condition is that J may not depend on the physical parameters, which also explains the name “structure matrix”. Choose a state x such that the electrical network of Example 2.1.2 can be written as a port-Hamiltonian system with ⎡ ⎤ 0 0 1 J =⎣ 0 0 1 ⎦. −1 −1 0 Determine the Hamiltonian density in this situation and show that the Hamiltonian remains unchanged.

2.5 Notes and references After the famous work of Kalman [29] in the 1960s, state space formulation has become one of the most used representation of systems. We refer to [25], [32], and [50] for further results on state space representations. Concerning the results on port-Hamiltonian systems we followed van der Schaft [55].

Chapter 3

Controllability of Finite-Dimensional Systems In this chapter we study the notion of controllability for finite-dimensional systems as introduced in Chapter 2. For this notion we only need the state differential equation x(t) ˙ = Ax(t) + Bu(t), x(0) = x0 , t ≥ 0, (3.1) where A and B are matrices, x0 is a vector and u is a locally integrable function. As before, we denote by K the set R or C and we denote the system (3.1) by Σ(A, B). We recall that the unique mild solution of (3.1) is given by At

t

eA(t−s) Bu(s) ds,

x(t) = e x0 +

t ≥ 0.

(3.2)

0

Intuitively, the concept of controllability concerns the problem of steering the state of the system from a given state into another state.

3.1 Controllability There are different notions of controllability available in the literature, most of them are equivalent for finite-dimensional systems. We will see later on, that this is not the case when it comes to infinite-dimensional systems. Definition 3.1.1. Let A ∈ Kn×n and B ∈ Kn×m . We call the system Σ(A, B) controllable, if for every x0 , x1 ∈ Kn there exists a t1 > 0 and a function u ∈ L1 ((0, t1 ); Km ) such that the mild solution x of (3.1), given by (3.2), satisfies x(t1 ) = x1 . A controllable system has the property that we are able to steer from every point to every other point in the state space. The ability to steer from the origin B. Jacob and H.J. Zwart, Linear Port-Hamiltonian Systems on Infinite-dimensional Spaces, Operator Theory: Advances and Applications 223, DOI 10.1007/978-3-0348-0399-1_3, © Springer Basel 2012

27

28

Chapter 3. Controllability of Finite-Dimensional Systems

to every point in the state space is known as reachability. This will be defined next. It is clear that controllability implies reachability. In fact, for the system (3.1) these notions are equivalent, see Theorem 3.1.6. Definition 3.1.2. The system Σ(A, B) is reachable if for every x1 ∈ Kn there exists a t1 > 0 and a function u ∈ L1 ((0, t1 ); Km ) such that the unique mild solution of (3.1) with x0 = 0 satisfies x(t1 ) = x1 . In order to characterize controllability of the system Σ(A, B) we introduce the controllability matrix and the controllability Gramian. Definition 3.1.3. Let A ∈ Kn×n and B ∈ Kn×m . We define the controllability matrix R(A, B) by

R(A, B) := B, AB, · · · , An−1 B . (3.3) The controllability Gramian Wt , t > 0, is defined by Wt :=

t

eAs BB ∗ eA

∗

s

ds.

(3.4)

0

It is clear that Wt is an n × n-matrix, whereas R(A, B) is an element of Kn×nm . For t > 0 it is easy to see that the matrix Wt is positive semi-definite, i.e. x∗ Wt x ≥ 0 for all x ∈ Kn . Moreover, Wt is positive definite if and only if Wt is invertible. In the following we frequently use the Theorem of Cayley-Hamilton, which we formulate next. For a proof we refer to standard textbooks on linear algebra. Theorem 3.1.4 (Theorem of Cayley-Hamilton). Let A ∈ Kn×n with the characteristic polynomial given by p(λ) := det(λI − A). Then A satisfies p(A) = 0. The Theorem of Cayley-Hamilton states that An can be expressed as a linear combination of the lower matrix powers of A. We need some more standard definitions from linear algebra. For a matrix T ∈ Kp×q , the rank of T is denoted by rk T , and the subspace ran T , the range of T , is defined by ran T := {y ∈ Kp | y = T x for some x ∈ Kq }. The range of the controllability Gramian Wt is independent of t and equals the range of the controllability matrix as it is shown in the following proposition. Proposition 3.1.5. For every t > 0 we have ran Wt = ran R(A, B). In particular, Wt is positive definite if and only if rk R(A, B) = n. Proof. We show that (ran Wt )⊥ = (ran R(A, B))⊥ . Let us first assume that x ∈ (ran R(A, B))⊥ . Thus x∗ Ak B = 0 for k = 0, . . . , n − 1. By the Theorem of Cayley-Hamilton we obtain x∗ Ak B = 0,

k ∈ N,

3.1. Controllability

29

and in particular x∗ eAs B =

∞ sk x∗ Ak B

k!

k=0

s ≥ 0.

= 0,

(3.5)

Let t > 0 be arbitrary. It follows from (3.5) that

∗

x Wt =

t

x∗ eAs BB ∗ eA

∗

s

ds = 0,

0

and therefore x ∈ (ran Wt )⊥ . Conversely, let x ∈ (ran Wt )⊥ for some t > 0. This implies ∗

t

0 = x Wt x =

∗

B ∗ eA s x 2 ds.

0 ∗

As the function s → B ∗ eA s x 2 is continuous and non-negative, we obtain ∗ B ∗ eA s x = 0 for every s ∈ [0, t]. In particular, x∗ B = 0. Moreover, we obtain dk ∗ A∗ s 0 = k (B e x) = B ∗ (A∗ )k x, ds s=0

k ∈ N.

This implies x∗ Ak B = 0 for every k ∈ N, and thus in particular x ∈ (ran R(A, B))⊥ . We are now in the position to characterize controllability. In particular, we show that if the system Σ(A, B) is controllable, then it is controllable in arbitrarily short time. We note, that this result is no longer true for infinite-dimensional systems. Theorem 3.1.6. Let A ∈ Kn×n , B ∈ Kn×m and t1 > 0. Then the following statements are equivalent: 1. For every t1 > 0 the system Σ(A, B) is controllable in time t1 , that is, for every x0 , x1 ∈ Kn there exists a function u ∈ L1 ((0, t1 ); Km ) such that the unique mild solution of (3.1) satisfies x(t1 ) = x1 . 2. The system Σ(A, B) is controllable. 3. The system Σ(A, B) is reachable. 4. The rank of R(A, B) equals n. Proof. Clearly part 1 implies part 2, and part 2 implies part 3. We now prove that part 3 implies part 4. Therefore, we assume that the system Σ(A, B) is reachable. Let x ∈ Kn be arbitrary. It is sufficient to show that

30


x ∈ ran R(A, B). As the system Σ(A, B) is reachable, there exists a time t1 > 0 and a function u ∈ L1 ((0, t1 ); Km ) such that t1 x= eA(t1 −s) Bu(s) ds. 0

Using Theorem 3.1.4 and the fact that ran R(A, B) is a (closed) subspace of Kn , we obtain e

A(t1 −s)

Thus

Bu(s) =

x= 0

∞ t1 k=0

∞ (t1 − s)k

Ak Bu(s) ∈ ran R(A, B).

k!

k=0

∈ran R(A,B)

(t1 − s)k k A Bu(s) ds ∈ ran R(A, B), k!

where we have used again the fact that every subspace of a finite-dimensional space is closed. Thus part 4 is proved. Finally, we prove the implication from part 4 to part 1. Thus we assume that rk R(A, B) = n. Let x ∈ Kn and t1 > 0 be arbitrary. We have to show that there is a function u ∈ L1 ((0, t1 ); Km ) such that t1 At1 x1 = e x0 + eA(t1 −s) Bu(s) ds. 0

By Proposition 3.1.5, there exists a vector y ∈ Kn such that t1 ∗ x1 − eAt1 x0 = Wt1 y = eAs BB ∗ eA s y ds.

(3.6)

0

We define the function u ∈ L1 ((0, t1 ); Km ) by u(s) := B ∗ eA

∗

(t1 −s)

Using (3.6) and (3.7), we obtain t1 At1 As ∗ A∗ s x1 − e x0 = e BB e y ds = 0

In other words, x1 = eAt1 x0 +

t1

e 0

t1 0

y,

s ∈ [0, t1 ].

(3.7)

As

t1

eA(t1 −s) Bu(s) ds.

Bu(t1 − s) ds = 0

eA(t1 −s) Bu(s) ds and thus the theorem is proved.

Theorem 3.1.6 shows that controllability can be checked by a simple rank condition. Furthermore, we see that if the system is controllable, then the control ∗ can be chosen as u(s) = B ∗ eA (t1 −s) Wt−1 (x1 − eAt1 x0 ), see (3.6) and (3.7). In 1 particular, the control can be chosen to be smooth, whereas in the definition we only required that the control is integrable. We close this session with some examples.

3.1. Controllability

31 s u M

θ m Figure 3.1: Cart with pendulum Example 3.1.7. We consider a cart on a container bridge to which a pendulum has been attached, see Figure 3.1. The pendulum represents the clamshell. The cart is driven by a motor which at time t exerts a force u(t) taken as control. We assume that all motion occurs in a plane, that is the cart moves along a straight line. Let M > 0 be the mass of the cart, m > 0 be the mass of the pendulum, which we assume is concentrated at the tip, > 0 be the length of the pendulum, s be the displacement of the center of the cart with respect to some fixed point, θ be the angle that the pendulum forms with the vertical, and g be the acceleration of gravity. We assume that the angle θ is small, and thus the kinectic energy equals 1 1 1 2 2 ˙ 2 2 M s˙ + 2 m(s˙ + θ) and the potential energy 2 mg θ . The Lagrange’s equations give (M + m)¨ s + m θ¨ = u, s¨ + θ¨ + gθ = 0. Thus we obtain the following linear system of differential equations mg 1 M s¨ = u + mgθ, θ¨ = − 1 + θ− u. M M ˙ T as state variable, the corresponding state space repreChoosing x := (s, s, ˙ θ, θ) sentation is given by ⎡ ⎡ ⎤ ⎤ 0 1 0 0 0 mg ⎢0 0 ⎢ 1 ⎥ 0⎥ M ⎥ x(t) + ⎢ M ⎥ u(t). x(t) ˙ =⎢ ⎣0 0 ⎣ 0 ⎦ 0 1⎦ m g 1 0 0 −(1 + M ) 0 − M =:A

=:B

Thus the controllability matrix R(A, B) of the system is given by ⎡ ⎤ mg 1 0 0 −M 2 M mg ⎢ 1 ⎥ 0 −M 0 2 M ⎥ R(A, B) = ⎢ g ⎦. 1 m ⎣ 0 − M 0 (1 + M ) M 2 g 1 m − M 0 (1 + M ) M 0 2

32


Dividing the first and the second row by and adding it to the third and fourth row, respectively, we find that R(A, B) is similar to ⎡

0

⎢1 ⎢M ⎣0 0

1 M

0 0 0

0 mg −M 2 0 g M2

⎤ mg −M 2 0 ⎥ ⎥. g ⎦ M2 0

As the rank of this matrix is 4, the rank of R(A, B) is 4, and thus the system is controllable. Example 3.1.8. Consider the electrical network given by Figure 3.2. Here V denotes the voltage source, L1 , L2 denote the inductance of the inductors, and C denotes the capacitance of the capacitor. As in Example 1.1.2, using Kirchhoff’s laws and

V L1

L2

C

Figure 3.2: Electrical network C the equations VL (t) = L dIdtL (t) and IC (t) = C dV dt (t), see (1.3) and (1.4), we obtain the system equations

dIL1 = VL1 = VC − V, (3.8) dt dIL2 L2 = VL2 = VC − V, and (3.9) dt dVC C = IC = −IL1 − IL2 . (3.10) dt x1 IL1 Choosing the state x = xx2 = IL2 and the control u(t) = V (t), we receive L1

3

VC

the state space representation ⎡

0 x(t) ˙ =⎣ 0 − C1

0 0 − C1

⎤ − L11 ⎦ x(t) + ⎣ − 1 ⎦ u(t). L2 0 0

1 L1 1 L2

⎤

⎡

(3.11)

3.2. Normal forms

33

The controllability matrix is given by ⎡ 0 − L11 ⎢ −1 0 R(A, B) = ⎣ L2 1 0 + CL1

1 CL2

1 CL21 1 CL22

⎤ + CL11 L2 + CL11 L2 ⎥ ⎦. 0

(3.12)

Since L1 times the first row equals L2 times the second row, we obtain rk R(A, B) = 2 and thus the system Σ(A, B) is not controllable. The lack of controllability can also be seen from equation (3.8) and (3.9). The voltage over the inductors is the same independent of the input u. Hence L1 x1 (t) − L2 x2 (t) is constant, and so we cannot choose x1 (t) and x2 (t) independent of each other. Their relation is determined by the initial state. We have seen in Chapter 2 that the choice for the state is to some extent arbitrary. For example, we can just interchange two state variables. However, the property of controllability is independent of this particular choice of state variables, see Lemma 3.2.2. Thus it is desirable to choose the state such that the matrices A and B have a special form. This question is studied in the next section.

3.2 Normal forms In the last paragraph of the previous section we discussed the fact that the state is non-unique. Therefore it is often useful to study systems under a change of basis in the state space Kn . Let T ∈ Kn×n be invertible. Then the basis transformation x ˆ := T −1 x transforms the system (3.1) into the system, see also Exercise 2.1, x ˆ˙ (t) = T −1 AT x ˆ(t) + T −1 Bu(t),

x ˆ(0) = T −1 x0 ,

t ≥ 0.

(3.13)

This leads to the following definition. ˆ ∈ Kn×m . The systems Σ(A, B) and Definition 3.2.1. Let A, Aˆ ∈ Kn×n and B, B ˆ ˆ Σ(A, B) are called similar, if there exists an invertible matrix T ∈ Kn×n such that ˆ B) ˆ = (T −1 AT, T −1B). (A,

(3.14)

It is easy to see that the controllability matrices of similar systems are related by ˆ B) ˆ = T −1 R(A, B), R(A,

(3.15)

ˆ B) ˆ are as in (3.14). This immediately leads to the following where (A, B) and (A, result. ˆ B) ˆ be as in (3.14). The system Σ(A, B) is conLemma 3.2.2. Let (A, B) and (A, ˆ B) ˆ is controllable. trollable if and only if the system Σ(A,

34


In particular, the statement of the lemma shows that controllability is a property of the system, and not of the particular representation we have chosen ˆ B) ˆ has a simple for the system. Thus we may try to find a matrix T for which (A, form or equivalently to find a normal form for the system Σ(A, B). An application of the following normal form will be given in the next chapter. Theorem 3.2.3. Let A ∈ Kn×n and b ∈ Kn×1 be such that Σ(A, b) is controllable. Then there exists an invertible matrix T ∈ Kn×n such that ⎡ ⎤ 0 1 0 ··· 0 ⎡ ⎤ 0 ⎢ .. ⎥ .. .. .. ⎢ 0 . . . . ⎥ ⎢ .. ⎥ ⎢ ⎥ ⎥ ˆb := T b = ⎢ ⎥. . . . ⎢ . ⎥ and Aˆ := T AT −1 = ⎢ . . . ⎢ . . 0 ⎥ ⎣0⎦ ⎢ . ⎥ ⎣ 0 ··· ··· 0 1 ⎦ 1 −a0 −a1 · · · · · · −an−1 The numbers a0 , . . . , an−1 ∈ K are uniquely determined !n−1as the coefficients of the characteristic polynomial of A, i.e., det(sI − A) = j=0 aj sj + sn . Proof. Let R := R(A, b). The assumption that Σ(A, b) is controllable implies that the matrix R is invertible. We write ⎡ ∗⎤ v1 ⎢ ⎥ R−1 = ⎣ ... ⎦ vn∗ with vi ∈ Kn , i = 1, . . . , n. By the definition of R and vn∗ we have " 0, j = 0, . . . , n − 2, vn∗ Aj b = 1, j = n − 1.

(3.16)

Next we show that the vectors vn∗ , vn∗ A, . . . , vn∗ An−1 are linearly independent. Therefore we assume that there exist scalars α1 , . . . , αn such that n

αi vn∗ Ai−1 = 0.

(3.17)

i=1

Multiplying this equation from the right by b yields n

αi vn∗ Ai−1 b = 0.

i=1

Thus (3.16) implies αn = 0. Multiplication of (3.17) by Ab and using again (3.16), then implies αn−1 = 0. In general, multiplying (3.17) sequently by b, Ab, A2 b, . . .

3.2. Normal forms

35

and using (3.16), implies α1 = · · · = αn = 0, that is, the vectors vn∗ , vn∗ A, . . . , vn∗ An−1 are linearly independent. In particular, this implies that the matrix ⎡ ⎤ vn∗ ⎢ vn∗ A ⎥ ⎢ ⎥ T := ⎢ (3.18) ⎥ ∈ Kn×n .. ⎣ ⎦ . vn∗ An−1

is invertible. By a0 , . . . , an−1 ∈ K we denote the coefficients of the characteristic polynomial of the matrix A, that is, det(sI − A) =

n−1

aj sj + sn .

j=0

The Theorem of Cayley-Hamilton now implies that An = − vn∗ An = −

n−1

aj vn∗ Aj .

!n−1 j=0

aj Aj and thus (3.19)

j=0

Combining (3.19), (3.18) and ⎡ 0 ⎡ ∗ ⎤ ⎢ vn A ⎢ 0 ⎢ .. ⎥ ⎢ TA = ⎣ . ⎦ = ⎢ ⎢ ... ⎢ vn∗ An ⎣ 0 −a0

(3.16), it is easy to see that ⎤ 1 0 ··· 0 ⎡ ⎤ 0 .. ⎥ .. .. .. ⎥ . . . . ⎥ ⎢ .. ⎥ ⎥ ⎥ T and T b = ⎢ .. .. ⎢.⎥ . . . 0 ⎥ ⎣ 0⎦ ⎥ ··· ··· 0 1 ⎦ 1 −a1 · · · · · · −an−1

This proves the assertion of the theorem.

The previous theorem shows that every controllable system Σ(A, b) is similar to a simple system, only containing ones, zeros, and the coefficients of the characteristic polynomial. The following result, known as Kalman controllability decomposition, is often useful in order to find simple proofs concerning controllability. Theorem 3.2.4. Let A ∈ Kn×n and B ∈ Kn×m . If the system Σ(A, B) is not controllable, that is, rk R(A, B) = r < n, then Σ(A, B) is similar to a system ˆ B) ˆ of the form Σ(A, B A1 A2 ˆ ˆ , 1 , (3.20) (A, B) = 0 A4 0 with suitable matrices A1 ∈ Kr×r , A2 ∈ Kr×(n−r) , A4 ∈ K(n−r)×(n−r) and B1 ∈ Kr×m . Moreover, the system Σ(A1 , B1 ) is controllable.

36


Proof. Let Z := ran R(A, B) = span {v1 , . . . , vr } ⊂ Kn . Thus we can find vectors vr+1 , . . . , vn ∈ Kn such that the vectors v1 , . . . , vn form a basis of Kn . We define

T := v1 , . . . , vn ∈ Kn×n . Therefore, T is invertible. We define ˆ := T −1 B B

and

Aˆ := T −1 AT.

ˆ B) ˆ are similar. As These definitions imply that the systems Σ(A, B) and Σ(A, ran B ⊂ Z, we see ˆ := T −1 B = B1 B 0 with B1 ∈ Kr×m . By the Theorem of Cayley-Hamilton we have ran AR(A, B) ⊂ ran R(A, B) = Z = span {v1 , . . . , vr } and thus A1 A2 ˆ AT = T A = T 0 A4 with suitable matrices A1 ∈ Kr×r , A2 ∈ Kr×(n−r) and A4 ∈ K(n−r)×(n−r) . Using ˆ we obtain the representations of Aˆ and B B1 A1 B1 · · · An−1 B1 1 ˆ ˆ R(A, B) = = T −1 R(A, B). 0 0 ··· 0 Since multiplication by an invertible matrix does not change the rank, the rank of ˆ B) ˆ equals r = rk R(A, B). Furthermore, we obtain R(A,

B = r. rk B1 , A1 B1 , . . . , An−1 1 Since A1 is an r × r matrix, the

Theorem of Cayley-Hamilton implies that the rank of B1 , A1 B1 , . . . , An−1 B equals the rank of R(A1 , B1 ). Thus the rank of 1 R(A1 , B1 ) equals the dimension of the associated state space, and thus Σ(A1 , B1 ) is controllable.

3.3 Exercises 3.1. In this exercise we obtain some additional tests for controllability of the system Σ(A, B) with A ∈ Cn×n and B ∈ Cn×m . (a) Show that the system Σ(A, B) is controllable if and only if for every eigenvector v of A∗ we have v ∗ B = 0. (b) Show that the system Σ(A, B) is controllable if and only if rk[sI − A, B] = n for all s ∈ C.

3.3. Exercises

37

3.2. Consider the second-order system z¨(t) = A0 z(t) + B0 u(t),

(3.21)

where z and u are vector-valued functions taking values in Kn and Km , respectively. In this exercise we write this as a state space model, and we investigate the controllability of this model in terms of A0 and B0 . z(t) to formulate (3.21) as the state (a) Use the state vector x(t) = z(t) ˙ differential equation x(t) ˙ = Ax(t) + Bu(t). (b) Prove that the state differential equation found in the previous part is controllable if and only if R(A0 , B0 ) has rank n. 3.3. We study a variation of Example 3.1.7, see Figure 3.3. Again we consider a cart on a container bridge, but now we attach two pendulums to the cart, see Figure 3.3. 1 > 0 and 2 > 0 are the lengths of the pendulums, and we assume that the mass of both pendulums are the same. As in Example 3.1.7 s u M

2 θ2

1 θ1

Figure 3.3: Cart with two pendulums the cart is driven by a motor which at time t exerts a force u(t) taken as control. Neglecting friction and assuming that the angles θ1 and θ2 are small, we obtain the following linear system of differential equations M s¨ = u + mg(θ1 + θ2 ), m gθ1 − 1 θ¨1 = − 1 + M m gθ2 − 2 θ¨2 = − 1 + M

mg θ2 − M mg θ1 − M

1 u, M 1 u. M

Use Exercise 3.2 to show that the system is controllable if and only if 1 = 2 . 3.4. We continue with the system discussed in Exercise 3.3.

38

Chapter 3. Controllability of Finite-Dimensional Systems (a) Assume that 1 = 2 . Find the Kalman controllability decomposition of the system, see Theorem 3.2.4. (b) Assume that 1 = 2 . Write the state space representation of the system in normal form, see Theorem 3.2.3.

3.4 Notes and references The results of this chapter can be found for example in [50] and [32]. The characterization of Exercise 3.1.b is known as the Hautus test, see [22].

Chapter 4

Stabilizability of Finite-Dimensional Systems This chapter is devoted to the stability and stabilizability of state differential equations. Roughly speaking, a system is stable if all solutions converge to zero, and a system is stabilizable if one can find a suitable control function such that the corresponding solution tends to zero. Thus stabilizability is a weaker notion than controllability.

4.1 Stability and stabilizability We begin by defining stability for the homogeneous state space equation and consider the system x(t) ˙ = Ax(t),

x(0) = x0 ,

t ≥ 0.

(4.1)

Here A is an n × n-matrix and x0 is an n-dimensional vector. As in the previous chapters we denote by K either R or C. Definition 4.1.1. Let A ∈ Kn×n . The differential equation (4.1) is called exponentially stable, if for every initial condition x0 ∈ Kn the solution of (4.1) satisfies lim x(t) = 0.

t→∞

In Chapter 2 we have shown that the unique solution of (4.1) is given by x(t) = eAt x0 , t ≥ 0. Using the Jordan normal form for square matrices, the exponential of A can be calculated as eAt = V eΛt V −1 , where V is an invertible matrix whose columns consist of the generalized eigenvectors of the matrix A, and Λ is a matrix which has non-zero entries only on the diagonal and on the superdiagonal. Moreover, the diagonal elements of the matrix Λ are equal to the B. Jacob and H.J. Zwart, Linear Port-Hamiltonian Systems on Infinite-dimensional Spaces, Operator Theory: Advances and Applications 223, DOI 10.1007/978-3-0348-0399-1_4, © Springer Basel 2012

39

40

Chapter 4. Stabilizability of Finite-Dimensional Systems

eigenvalues of A. Thus the differential equation (4.1) is exponentially stable if and only if all eigenvalues of the matrix A lie in the open left half-plane of C. A matrix with the property that all its eigenvalues lie in the left half-plane is called a Hurwitz matrix. In the following we denote by σ(A) the set of all eigenvalues of the matrix A. From the representation of the exponential of a matrix, the following result follows immediately. Lemma 4.1.2. The differential equation (4.1) is exponentially stable if and only if there exist constants M ≥ 1 and ω > 0 such that

eAt x0 ≤ M e−ωt x0 ,

t ≥ 0, x0 ∈ Kn .

Thus, if all solutions of (4.1) converge to zero for t → ∞, then all solutions converge uniformly and exponentially to zero. This motivates the notion “exponentially stable”. If the matrix A possesses eigenvalues with non-negative real part, then there are solutions of equation (4.1) which do not tend to zero. It is thus desirable to stabilize the system, that is, we study again a system with an input x(t) ˙ = Ax(t) + Bu(t),

x(0) = x0 ,

t ≥ 0,

(4.2)

and we try to find a suitable input function u such that the corresponding solution x converges to zero for t → ∞. Definition 4.1.3. Let A ∈ Kn×n and B ∈ Kn×m . We call the system Σ(A, B) stabilizable, if for every x0 ∈ Kn there exists a function u ∈ L1loc ((0, ∞); Km ) such that the unique solution of (4.2) converges to zero for t → ∞. Clearly, controllable systems Σ(A, B) are stabilizable. We only have to steer the state to zero in some time t1 and to choose the input u to be zero on the interval [t1 , ∞), see Exercise 4.1.

4.2 The pole placement problem It is the aim of the following two sections to characterize stabilizability and to show that the stabilizing control function u can be obtained via a feedback law u(t) = F x(t). In order to show the existence of such a feedback, we first have to study the more general problem of pole placement: Pole placement problem. Given A ∈ Kn×n , B ∈ Kn×m and complex numbers λ1 , . . . , λn the question is whether there exists a matrix F ∈ Km×n such that λ1 , . . . , λn are the eigenvalues of the matrix A + BF . If the pole placement problem is solvable, then we can move the eigenvalues of closed loop system x(t) ˙ = Ax(t) + Bu(t),

u(t) = F x(t),

t ≥ 0,

(4.3)

4.2. The pole placement problem

41

to arbitrary points in C. Note that equations (4.3) are equivalent to x(t) ˙ = (A + BF )x(t)

t ≥ 0.

So far all results hold for real- and complex-valued matrices. However, the situation is different for the pole placement problem. For example, the points λ1 , . . . , λn are always allowed to be complex. To simplify the exposition we formulate results for the complex situation only. In the real situation the results hold with obvious modifications. The following theorem shows that the system Σ(A, B) is controllable if the pole placement problem is solvable. Theorem 4.2.1. Let A ∈ Cn×n and B ∈ Cn×m . Then 1. implies 2., and 2. implies 3., where 1. The pole placement problem is solvable. 2. There exists an F ∈ Cm×n such that the eigenvalues of A + BF are all different from the eigenvalues of A, i.e., σ(A + BF ) ∩ σ(A) = ∅. 3. The system Σ(A, B) is controllable. We note that the three statements of Theorem 4.2.1 are equivalent, see Corollary 4.2.6. Proof. Clearly 1. implies 2. Thus we concentrate on the implication from 2. to 3. If Σ(A, B) is not controllable, then rk R(A, B) =: r < n. By Theorem 3.2.4 there exists an invertible T such that A1 A2 B , 1 (T −1 AT, T −1 B) := 0 A4 0 with suitable matrices A1 ∈ Cr×r , A2 ∈ Cr×(n−r) , A4 ∈ C(n−r)×(n−r) and B1 ∈ Cr×m . If we write an arbitrary matrix F ∈ Cm×n in the form F = F1 F2 T −1 with F1 ∈ Cm×r , we obtain A1 + B1 F1 A2 + B1 F2 −1 A + BF = T T . 0 A4 As the eigenvalue of A4 are fixed, the intersection of the eigenvalues of A+BF and A always contains the eigenvalues of A4 . Thus for every F this intersection is non empty, which contradicts the assertion in 2. Concluding, Σ(A, B) is controllable. In the following we show, that controllability is actually equivalent to the solvability of the pole placement problem. Theorem 4.2.2. Let A ∈ Cn×n , b ∈ Cn and λ1 , . . . , λn ∈ C. If the system Σ(A, b) is controllable, then there exists a matrix f ∈ C1×n such that the matrix A + bf has the eigenvalues λ1 , . . . , λn , counted with multiplicities.

42


Proof. By Theorem 3.2.3 there exists an invertible matrix ⎡ 0 1 0 ⎡ ⎤ 0 ⎢ . . .. .. ⎢ 0 ⎢ .. ⎥ ⎢ ⎥ −1 . ˆb := T b = ⎢ ˆ ⎢ .. A := T AT = ⎢ .. ⎢ ⎥, . ⎣0⎦ ⎢ . ⎣ 0 ··· ··· 1 −a0 −a1 · · · and det(sI − A) =

n−1

T ∈ Cn×n such that ⎤ ··· 0 .. ⎥ .. . . ⎥ ⎥ ⎥ .. (4.4) . 0 ⎥ ⎥ 0 1 ⎦ · · · −an−1

aj sj + sn .

j=0

We define the polynomial p by p(s) =

n #

(s − λj ) = sn +

j=1

n−1

p j sj ,

(4.5)

and f = fˆT.

(4.6)

j=0

and the matrices fˆ, f ∈ C1×n by

fˆ := a0 − p0 , . . . , an−1 − pn−1

Thus A + bf = T −1 (Aˆ + ˆbfˆ)T , which implies that the set of eigenvalues of A + bf equals the set of eigenvalues of Aˆ + ˆbfˆ, counted with multiplicities. Moreover, from (4.4) and (4.6) we obtain that ⎡ ⎤ 0 1 0 ··· 0 ⎢ .. ⎥ .. .. .. ⎢ 0 . . . . ⎥ ⎢ ⎥ ˆ ˆ ˆ ⎢ ⎥. . . . A + bf = ⎢ . . . . . 0 ⎥ ⎢ . ⎥ ⎣ 0 ··· ··· 0 1 ⎦ −p0 −p1 · · · · · · −pn−1 Thus the characteristic polynomial of Aˆ + ˆbfˆ equals p. By (4.5) we conclude that Aˆ + ˆbfˆ and thus A + bf has exactly the eigenvalues λ1 , . . . , λn , again counted with multiplicities. Proposition 4.2.3. Let A ∈ Cn×n and B ∈ Cn×m be such that Σ(A, B) is controllable and let b ∈ Cn \{0}. Then there exists a matrix F ∈ Cm×n such that the system Σ(A + BF, b) is controllable. Proof. We define r := rk R(A, b) ≤ n. If r = n, then the system Σ(A, b) is controllable and thus we can choose F = 0. Let r < n. By Theorem 3.2.4, we can assume without loss of generality that b A1 A2 , 1 (A, b) = 0 A4 0

4.2. The pole placement problem

43

with suitable matrices A1 ∈ Cr×r , A2 ∈ Cr×(n−r) , A4 ∈ C(n−r)×(n−r) , b1 ∈ Cr , and Σ(A1 , b1 ) is controllable. Since Σ(A1 , b1 ) is controllable, the square matrix b1 ] [b1 , A1 b1 , · · · , Ar−1 1 has rank r. Thus by the Theorem of Cayley-Hamilton b A1 b1 · · · Ar−1 b A1 b1 · · · b1 1 r = rk 1 = rk 1 0 0 ··· 0 0 0 ···

An−1 b1 1 = rk R(A, b). 0 (4.7) This implies that the vectors vi := Ai−1 b, i = 1, . . . , r, are linearly independent and span {v1 , . . . , vr } = ran R(A, b). As the system Σ(A, B) is controllable, we have rk R(A, B) = n. Next we want to show that there exists a vector ˆb ∈ ran B\ran R(A, b). We assume that this does not hold, that is, we assume that ran B ⊂ ran R(A, b). Then the Theorem of Cayley-Hamilton implies ran R(A, B) ⊂ ran R(A, b), which is in contradiction to rk R(A, b) = r < n = rk R(A, B). Let ˆb ∈ ran B\ran R(A, b) and u ∈ Cm with Bu = ˆb. We define vr+1 = Avr + ˆb. Since by (4.7) Avr ∈ ran R(A, b) = span {v1 , . . . , vr }, this choice for vr+1 implies that the vectors v1 , . . . , vr+1 are linearly independent. We choose a matrix F ∈ Cn×n with the property " 0, i = 1, . . . , r − 1, F vi = u, i = r, and we define F1 := A + BF. Then we have F1i−1 b = vi ,

i = 1, . . . , r + 1,

and thus rk R(F1 , b) ≥ r + 1. If rk R(F1 , b) = n, then the statement of the proposition follows. If rk R(F1 , b) < n, then we first show that Σ(F1 , B) is controllable. By Theorem 3.1.6 it is sufficient to prove that Σ(F1 , B) is reachable. Let x1 ∈ Cn be arbitrarily. Since Σ(A, B) is reachable, there is some t1 > 0 and an input u ∈ L1 ((0, t1 ); Cm ) such that the mild solution of x(t) ˙ = Ax(t) + Bu(t), x(0) = 0, satisfies x(t1 ) = x1 . Now we choose v = u − F x as input for the system Σ(F1 , B) and the mild solution of x(t) ˙ = F1 x(t) + Bv(t), x(0) = 0, satisfies x(t1 ) = x1 , which shows that Σ(F1 , B) is reachable. As Σ(F1 , B) is controllable as well, we can apply the above procedure to the system Σ(F1 , b), until the rank of the matrix R(Fj , b) equals n. Next we are in the position to show that controllability implies the solvability of the pole placement problem.

44


Theorem 4.2.4. Let A ∈ Cn×n and B ∈ Cn×m be such that the system Σ(A, B) is controllable and let λ1 , . . . , λn ∈ C. Then there exists a matrix F ∈ Cm×n such that the matrix A + BF has exactly the eigenvalues λ1 , . . . , λn , counted with multiplicities. Proof. We choose a vector b = Bu ∈ ran B\{0}. By Proposition 4.2.3 there exists a matrix Fˆ ∈ Cm×n such that the system Σ(A + B Fˆ , b) is controllable. Finally, by Theorem 4.2.2 there exists a vector f such that the matrix A + B Fˆ + bf has exactly the eigenvalues λ1 , . . . , λn , counted with multiplicities. Thus, by the choice of F := Fˆ + uf the statement of the theorem is proved. As a consequence of the theorem, we obtain that for every controllable system Σ(A, B) there exists a matrix F ∈ Rm×n such that A + BF is a Hurwitz matrix. Corollary 4.2.5. Let A ∈ Cn×n and B ∈ Cn×m . If the system Σ(A, B) is controllable, then there exists a matrix F ∈ Rm×n such that A + BF is a Hurwitz matrix. Moreover, we have shown the equivalence of the pole placement problem and controllability. Corollary 4.2.6. Let A ∈ Cn×n and B ∈ Cn×m . Then the following statements are equivalent: 1. The pole placement problem is solvable. 2. There exists an F ∈ Cm×n such that the eigenvalues of A + BF are all different from the eigenvalues of A, i.e., σ(A + BF ) ∩ σ(A) = ∅. 3. The system Σ(A, B) is controllable. If we want to place the poles arbitrarily, then we need controllability. However, if one only wants to stabilize the system, then a weaker condition suffices as shown in the following section.

4.3 Characterization of stabilizability In Definition 4.1.3 we have defined stabilizability as an open loop notion, i.e., there exists an input such that the corresponding state converges to zero. In this section we show that if it is possible to stabilize the system via a well-chosen input, then it is also possible via state feedback, i.e., via an input of the form u(t) = F x(t). The following theorem formulates this assertion. Theorem 4.3.1. Let A ∈ Cn×n and B ∈ Cn×m . Then the following statements are equivalent: 1. The system Σ(A, B) is stabilizable. 2. There exists a matrix F ∈ Cm×n such that A + BF is a Hurwitz matrix.

4.3. Characterization of stabilizability

45

3. For every eigenvector v of A∗ which belongs to an eigenvalue λ ∈ C with Re λ ≥ 0 we have: v ∗ B = 0. Proof. Clearly, part 2. implies part 1. Next we show that part 1. implies part 3. In order to prove this implication we assume that the system Σ(A, B) is stabilizable and that there exists an eigenvector v of A∗ which belongs to an eigenvalue λ ∈ C with Re λ ≥ 0 such that v ∗ B = 0. It then follows that ¯n v ∗ B = 0, v ∗ An B = ((An )∗ v)∗ B = λ and thus

v ∗ eAt B = 0,

n ∈ N,

t ≥ 0.

As the system Σ(A, B) is stabilizable, there exists a function u ∈ such that the function t At eA(t−s) Bu(s) ds, t ≥ 0, x(t) := e v +

(4.8) L1loc ((0, ∞); Cm )

(4.9)

0

converges to zero for t → ∞. Multiplying equation (4.9) from the left by v ∗ we obtain t ¯ ¯ v ∗ eA(t−s) Bu(s) ds = v ∗ eλt v + 0 = eλt v 2 , t ≥ 0. v ∗ x(t) = v ∗ eAt v + 0

The left-hand side converges to zero for t → ∞, whereas the right-hand side does not converge to zero. This leads to a contradiction. Thus part 1. implies part 3. It remains to show that part 3. implies part 2. If the system Σ(A, B) is controllable, then the statement follows from Corollary 4.2.5. If the system Σ(A, B) is not controllable, then rk R(A, B) =: r < n. By Theorem 3.2.4 there exists an invertible square matrix T such that ˆ T −1 B) ˆ = T −1 A1 A2 T, T −1 B1 , (4.10) (A, B) = (T −1 AT, 0 A4 0 where A4 is a (n − r) × (n − r)-matrix and the system Σ(A1 , B1 ) is controllable. Next we show that all eigenvalues of A4 lie in the open left half plane. Let α be an eigenvalue of A∗4 and let v 2 be a corresponding eigenvector. Using (4.10) it is easy to see that v := T ∗ v02 ∈ Cn is an eigenvector of A∗ with respect to the eigenvalue α. This implies that ∗

B1 0 B T T −1 1 = 0 v2∗ = 0. v∗ B = v2 0 0 By assumption this can only happen if Re α < 0. This shows that A4 is a Hurwitz matrix. As the system Σ(A1 , B1 ) is controllable, by Corollary 4.2.5 there exists

46


a matrix F1 ∈ Cm×r such that A1 + B1 F1 is a Hurwitz matrix. Choosing the feedback F := F1 0 T , we obtain that

A A2 B A + B1 F1 A2 A + BF = T −1 1 T + T −1 1 F1 0 T = T −1 1 T. 0 A4 0 0 A4

1 F1 A2 Since A1 +B 0 A4 is a Hurwitz matrix, the matrix A + BF is Hurwitz as well. Thus part 3. implies part 2. Summarizing, we have shown that a stabilizable system can be stabilized by state feedback. However, we still lack a simple condition which characterizes stabilizability. This will be given next, for which we need the following concept. Definition 4.3.2. Let A ∈ Kn×n . We call a subspace Z ⊂ Kn A-invariant, if AZ ⊂ Z. Theorem 4.3.3. Let A ∈ Cn×n and B ∈ Cn×m . Then the following statements are equivalent: 1. The system Σ(A, B) is stabilizable. 2. There exist two A-invariant subspaces Xs and Xu of Cn such that a) Cn = Xs ⊕ Xu . b) The system Σ(A, B) is similar to Bs As 0 , . Σ Bu 0 Au c) The matrix As is a Hurwitz matrix. d) The system Σ(Au , Bu ) is controllable. Proof. We first prove the implication 2. ⇒ 1. Clearly, the system Σ(As , Bs ) is stabilizable by a matrix F = 0, and the system Σ(Au , Bu ) is stabilizable by Corollary 4.2.5. Thus it is easy to see that the system Σ(A, B) is stabilizable. Next we assume that Σ(A, B) is stabilizable. We now define Xs by the span of all generalized eigenspaces corresponding to eigenvalues of A with negative real part and Xu by the span of all generalized eigenspaces corresponding to eigenvalues of A with non-negative real part. Thus, Xs and Xu are A-invariant, Xs ∩Xu = {0} and Xs ⊕ Xu = Cn . If we choose our basis of Cn accordingly, it is easy to see that the system Σ(A, B) is similar to a system of the form As 0 Bs Σ , . 0 Au Bu By definition of the space Xs , the matrix As is a Hurwitz matrix. It remains to show that Σ(Au , Bu ) is controllable. We know that the system Σ(Au , Bu ) is stabilizable and that all eigenvalues of Au lie in the closed right half plane. Thus by Corollary 4.2.6 the system Σ(Au , Bu ) is controllable.

4.4. Stabilization of port-Hamiltonian systems

47

4.4 Stabilization of port-Hamiltonian systems In this section we return to the class of (finite-dimensional) port-Hamiltonian systems as introduced in Section 2.3. We show that port-Hamiltonian systems are stabilizable if and only if they are controllable. Furthermore, if such a system is stabilizable, then the system is stabilizable by (static) output feedback, i.e., by choosing the input u = −ky, with k > 0. We consider the port-Hamiltonian system from Section 2.3 which is given by x(t) ˙ = JHx(t) + Bu(t), ∗

y(t) = B Hx(t),

(4.11) (4.12)

where J is skew-adjoint, i.e., J ∗ = −J and H is a positive-definite matrix. Theorem 4.4.1. The system (4.11)–(4.12) is stabilizable if and only if it is controllable. Furthermore, it is controllable if and only if it is stabilizable by applying the feedback u(t) = −ky(t) = −kB ∗ Hx(t) with k > 0. Proof. We define a new norm · H on Cn by

x 2H :=

1 ∗ x Hx, 2

(4.13)

which is equivalent to the Euclidean norm, and along solutions we have d x(t) 2H = Re (u(t)∗ y(t)) . dt

(4.14)

For u ≡ 0 and for any initial condition x(0), this implies

x(t) H = x(0) H .

(4.15)

Thus all eigenvalues of the matrix JH lie on the imaginary axis. By Theorem 4.3.3 we conclude that the system is stabilizable if and only if it controllable. It remains to construct a stabilizing controller. So we assume that the system is controllable, and as a candidate for a stabilizing controller we choose u(t) = −ky(t) = −kB ∗ Hx(t) with k > 0. Using (4.14), we have d x(t) 2H = −k y(t) 2 . (4.16) dt Applying the feedback u(t) = −kB ∗ Hx(t), we obtain the state equation x(t) ˙ = (JH − kBB ∗ H) x(t).

(4.17)

Equation (4.16) shows that for every initial condition the function t → x(t) 2H is non-increasing, and this implies that all eigenvalues of JH − kBB ∗ H lie in the closed left half-plane. Hence we can conclude stability if there are no eigenvalues on the imaginary axis.

48


Let λ ∈ iR be a purely imaginary eigenvalue of JH−kBB ∗ H with eigenvector v ∈ Cn , i.e., (JH − kBB ∗ H)v = λv. (4.18) Multiplying this equation from the left by v ∗ H gives v ∗ HJHv − kv ∗ HBB ∗ Hv = λv ∗ Hv.

(4.19)

Multiplying the Hermitian conjugate of equation (4.18) from the right by Hv, and using the fact that H is self-adjoint and J is skew-adjoint, we obtain −v ∗ HJHv − kv ∗ HBB ∗ Hv = −λv ∗ Hv. Adding this equation to equation (4.18), we find that kv ∗ HBB ∗ Hv = 0. Since k > 0, this is equivalent to (4.20) B ∗ Hv = 0. Equation (4.18) now implies that JHv = λv. Taking the Hermitian conjugate, we find λ∗ v ∗ = −λv ∗ = v ∗ H∗ J ∗ = −v ∗ HJ.

(4.21)

Combining (4.20) and (4.21), we see that v ∗ HJHB = −λ∗ v ∗ HB = λv ∗ HB = 0. Similarly, we find for k ∈ N, v ∗ H (JH)k B = 0,

(4.22)

which implies n−1 v ∗ HR(JH, B) = v ∗ H B, JHB, . . . , (JH) B = 0. As the system Σ(JH, B) is controllable, we obtain v ∗ H = 0, or equivalently, v = 0. This is in contradiction to the fact that v is an eigenvector. Summarizing, if the system is controllable, then the closed loop system matrix JH − kBB ∗ H has no eigenvalues in the closed right half-plane, and so we have stabilized the system.

4.5 Exercises 4.1. Assume that the system Σ(A, B) is controllable. Show that Σ(A, B) is also stabilizable by steering the state to zero in some time t1 and choosing the input u to be zero on the interval [t1 , ∞).


49

4.2. Show that the system Σ(A, B) is stabilizable if and only if

rk sI − A B = n for all s ∈ C with Re s ≥ 0. 4.3. In this exercise we study the system ⎡ ⎤ ⎡ ⎤ −2 0 3 1 x(t) ˙ = ⎣ 0 −1 0 ⎦ x(t) + ⎣0⎦ u(t). 3 0 −2 1 =:A

=:B

(a) Show that the system Σ(A, B) is not controllable. (b) Show that the system Σ(A, B) is stabilizable. (c) Find a stabilizing feedback control for this system. 4.4. Consider the following matrix ⎡

⎤ 0 1 6 1 ⎢ 1 0 1 7 ⎥ ⎥. A=⎢ ⎣ −4 0 0 −1 ⎦ 0 −5 −1 0

(4.23)

(a) Find a skew-adjoint J and a positive-definite matrix H such that A = JH. Hint: Assume J ∗ = J −1 and calculate A∗ A.

T (b) Now we choose b = 1 0 0 0 . Show that Σ(A, b) is controllable. (c) Design a stabilizing controller.

4.6 Notes and references Since stabilizability is one of the important problem within systems theory, its solution can be found in any book on finite-dimensional systems theory. We refer to our standard references [50] and [32]. The characterization of stabilizability for port-Hamiltonian systems can be found in [55]. The characterization of stabilizability of Exercise 4.2 is known as the Hautus test for stabilizability, see also Exercise 3.1.b.

Chapter 5

Strongly Continuous Semigroups In Chapter 2 we showed that the examples of Chapter 1, which were described by ordinary differential equations, can be written as a first-order differential equation x(t) ˙ = Ax(t) + Bu(t),

y(t) = Cx(t) = Du(t),

(5.1)

where x(t) is a vector in Rn or Cn . In Chapters 5 and 6 we show how systems described by partial differential equations can be written in the same form by using an infinite-dimensional state space. The formulation of the inputs and outputs is postponed till Chapter 10. Note that for partial differential equations the question of existence and uniqueness of solutions is more difficult than for ordinary differential equations. Thus we focus first on homogeneous partial differential equations. We begin by introducing the solution operator, and only in Section 5.3 do we show how to rewrite a p.d.e. as an abstract differential equation x(t) ˙ = Ax(t).

5.1 Strongly continuous semigroups We start with a simple example. Example 5.1.1. Consider a metal bar of length 1 that is insulated at the boundary: ∂x ∂2x (ζ, t) = (ζ, t) x(ζ, 0) = x0 (ζ), ∂t ∂ζ 2 ∂x ∂x (0, t) = 0 = (1, t). ∂ζ ∂ζ

(5.2)

x(ζ, t) represents the temperature at position ζ ∈ [0, 1] at time t ≥ 0 and x0 represents the initial temperature profile. The two boundary conditions state that B. Jacob and H.J. Zwart, Linear Port-Hamiltonian Systems on Infinite-dimensional Spaces, Operator Theory: Advances and Applications 223, DOI 10.1007/978-3-0348-0399-1_5, © Springer Basel 2012

51

52

Chapter 5. Strongly Continuous Semigroups

there is no heat flow at the boundary, see Example 1.1.5, and thus the bar is insulated. In order to calculate the solution of (5.2) we try to find a solution of the form x(ζ, t) = f (t)g(ζ). Substituting this equation in (5.2) and using the boundary conditions we obtain f (t)g(ζ) = αn e−n

2

π2 t

cos(nπζ),

(5.3)

where αn ∈ R or C, and n ∈ N. This solution satisfies the p.d.e. and the boundary conditions, but most likely not the initial condition. By the linearity of the p.d.e. (5.2) it is easy to see that xN (ζ, t) =

N

αn e−n

2

π2 t

cos(nπζ)

(5.4)

n=0

satisfies the p.d.e. and!the boundary conditions as well. The corresponding initial N condition xN (ζ, 0) = n=0 an cos(nπζ) is a Fourier polynomial. Note that every function q in L2 (0, 1) can be represented by its Fourier series, i.e., q(·) =

∞

αn cos(nπ·),

(5.5)

n=0 1

with equality/convergence in L2 (0, 1), α0 = 0 q(ζ) dζ and 1 q(ζ) cos(nπζ) dζ, n = 1, 2, . . . . αn = 2 0

If x0 ∈ L (0, 1), then we can define αn , n ∈ N ∪ {0} as the corresponding Fourier coefficients and ∞ 2 2 αn e−n π t cos(nπζ). (5.6) x(ζ, t) := 2

n=0 −n2 π 2 t

Since for t ≥ 0 we have e ≤ 1, the function x(·, t) is an element of L2 (0, 1). By construction, the initial condition is satisfied as well. However, as interchanging differentiation and (infinite) summation is not always allowed, it is unclear if this function satisfies the p.d.e. (5.2). Nevertheless the mapping x0 → x(·, t) defines an operator, which would assign to an initial condition its corresponding solution at time t, provided x is the solution. This example motivates the necessity for generalizing the concept of “eAt ” on abstract spaces and the necessity for clarifying the concept of “solution” of differential equations on abstract spaces. The answer is, of course, the well-known semigroup theory that we develop here for the special case of strongly continuous semigroups on a Hilbert space. We denote by X a real or complex (separable) Hilbert space, with inner $ product ·, · X and norm · X = ·, · X . By L(X) we denote the class of linear bounded operators from X to X.

5.1. Strongly continuous semigroups

53

Definition 5.1.2. Let X be a Hilbert space. (T (t))t≥0 is called a strongly continuous semigroup (or short C0 -semigroup) if the following holds: 1. For all t ≥ 0, T (t) is a bounded linear operator on X, i.e., T (t) ∈ L(X); 2. T (0) = I; 3. T (t + τ ) = T (t)T (τ ) for all t, τ ≥ 0; 4. For all x0 ∈ X, we have that T (t)x0 − x0 X converges to zero, when t ↓ 0, i.e., t → T (t) is strongly continuous at zero. We call X the state space, and its elements states. The easiest example of a strongly continuous semigroup is the exponential of a matrix. That is, let A be an n × n matrix, the matrix-valued function T (t) = eAt defines a C0 -semigroup on the Hilbert space Rn , see Exercise 5.1. Another example is presented next. Example 5.1.3. Let {φn , n ≥ 1} be an orthonormal basis of the separable Hilbert space X, and let {λn , n ≥ 1} be a sequence of complex numbers. Then ∞

T (t)x =

eλn t x, φn φn

(5.7)

n=1

is a bounded, linear operator if and only if {eRe λn t , n ≥ 1} is a bounded sequence in R, and this is the case for t > 0 if and only if sup Re λn < ∞. n≥1

Under this assumption, we have

T (t) ≤ eωt

(5.8)

with ω = supn≥1 Re λn . Furthermore, T (t + s)x =

∞

eλn (t+s) x, φn φn

n=1

and T (t)T (s)x = =

∞ n=1 ∞ n=1

eλn t T (s)x, φn φn =

∞

% eλn t

n=1

eλn t eλn s x, φn φn = T (t + s)x.

∞ m=1

& eλm s φm x, φm , φn

φn

54


Clearly, T (0) = I, and the strong continuity follows from the following calculation: For t ≤ 1 we have

T (t)x − x 2 =

∞

|eλn t − 1|2 |x, φn |2

n=1

=

N

|eλn t − 1|2 |x, φn |2 +

n=1

∞

|eλn t − 1|2 |x, φn |2

n=N +1 N

≤ sup |eλn t − 1|2 1≤n≤N

∞

|x, φn |2 + K

n=1

|x, φn |2

n=N +1

for K = sup0≤t≤1,n≥1 |eλn t − 1|2 . For any ε > 0 there exists an N ∈ N such that ∞

|x, φn |2
0 t

1 t

t 0

T (s)xds → x as t ↓ 0;

log T (t) ), then ω0 = lim ( 1t log T (t) ) < ∞; t→∞

5. For every ω > ω0 , there exists a constant Mω such that for every t ≥ 0 we have T (t) ≤ Mω eωt . The constant ω0 is called the growth bound of the semigroup. Proof. 1. First we show that T (t) is bounded on some neighborhood of the origin, that is, there exist δ > 0 and M > 1 depending on δ such that

T (t) ≤ M

for t ∈ [0, δ].

If this does not hold, then there exists a sequence {tn }, tn ↓ 0 such that T (tn ) ≥ n. Hence, by the uniform boundedness principle, there exists an element x ∈ X such that { T (tn )x , n ∈ N} is unbounded; but this contradicts the strong continuity at the origin. If we set t = mδ + τ with 0 < τ ≤ δ, then

T (t) = T (mδ)T (τ ) = T (δ)m T (τ )

≤ T (δ) m T (τ ) ≤ M 1+m ≤ M M t/δ = M eωt , where ω = δ −1 log M . 2. For fixed x ∈ X, t > 0 and s ≥ 0, using inequality (5.9) we have

T (t + s)x − T (t)x ≤ T (t)

T (s)x − x ≤ M eωt T (s)x − x . Hence by the strong continuity of T (·) we may conclude that lim T (t + s)x − T (t)x = 0. s↓0

Moreover, for x ∈ X, t > 0 and τ ≥ 0 sufficiently small, we have

T (t − τ )x − T (t)x ≤ T (t − τ )

x − T (τ )x .

(5.9)

56


Thus lim T (t + s)x − T (t)x = 0, and the mapping t → T (t)x is continuous on s↑0

the interval [0, ∞). 3. Let x ∈ X and ε > 0. By the strong continuity of (T (t))t≥0 we can choose τ > 0 such that T (s)x − x ≤ ε for all s ∈ [0, τ ]. For t ∈ [0, τ ] we have that 1 t 1 t T (s)x ds − x =

[T (s)x − x]ds

t 0 t 0 1 t 1 t ≤

T (s)x − x ds ≤ εds = ε. t 0 t 0 4. Let t0 > 0 be a fixed number and M = sup T (t) . For every t ≥ t0 there t∈[0,t0 ]

exists n ∈ N such that nt0 ≤ t < (n + 1)t0 . Consequently, log T (nt0 )T (t − nt0 )

log T n(t0 )T (t − nt0 )

log T (t)

= = t t t n log T (t0 ) log M ≤ + t t log T (t0 ) nt0 log M = · + . t0 t t log T (t0 ) + logtM if log T (t0 ) is positive, t0 log T (t0 ) t−t0 log M if log T (t0 ) is negative. Thus t0 t + t

The latter term is less than or equal to and it is less than or equal to lim sup t→∞

log T (t0 )

log T (t)

≤ < ∞, t t0

and since t0 is arbitrary, we have that lim sup t→∞

log T (t)

log T (t)

log T (t)

≤ inf ≤ lim inf . t→∞ t>0 t t t

Thus

log T (t)

log T (t)

= lim < ∞. t→∞ t t 5. If ω > ω0 , then there exists a t0 > 0 such that ω0 = inf

t>0

log T (t)

0: x0 − T (t)x0 T (−t)x0 − x0 + A+ x0 = T (−t) + A+ x0 + A+ x0 − T (−t)A+ x0 . t t (6.19) Since x0 ∈ D(A+ ) and (T (t))t∈R is strongly continuous, the right-hand side converges to zero for t ↓ 0, and thus also the left-hand side converges to zero for t ↓ 0. This implies that x0 ∈ D(A− ) and also that (6.15) exists, which shows D(A+ ) ⊂ D(A− ) and D(A+ ) ⊂ D(A). Since we already had that D(A) ⊂ D(A+ ), we conclude that D(A+ ) = D(A). Similarly, we can show that D(A− ) = D(A). Furthermore, from (6.19) we conclude that A− x0 = −Ax0 . Combining this with (6.18), we have that A+ = A, and A− = −A. Thus both A and −A are infinitesimal generators of C0 -semigroups. Next we prove the other implication. Denote the semigroups generated by A and −A by (T+ (t))t≥0 and (T− (t))t≥0 , respectively. For x0 ∈ D(A) consider the continuous function f (t) = T+ (t)T− (t)x0 , t ≥ 0. By Theorem 5.2.2 this function is differentiable, and df (t) = T+ (t)AT− (t)x0 + T+ (t)(−A)T− (t)x0 = 0. dt Thus f (t) = f (0) = x0 . Since T+ (t)T− (t) is a bounded operator, and since the domain of A is dense we have that T+ (t)T− (t) = I,

t ≥ 0.

(6.20)

Similarly, we find that T− (t)T+ (t) = I. Now we define (T (t))t∈R by (6.17). It is easy to see that (T (t))t∈R satisfies properties 1, 2, and 4 of Definition 6.2.1. So it remains to show that property 3 holds. We prove this for τ < 0, t > 0, and τ < −t. The other cases are shown similarly. We have that T (t + τ ) = T− (−t − τ ) = T+ (t)T− (t)T− (−t − τ ) = T+ (t)T− (−τ ) = T (t)T (τ ), where we have used (6.20) and the semigroup property of (T− (t))t≥0 .

Definition 6.2.4. A strongly continuous group (T (t))t∈R is called a unitary group if T (t)x = x for every x ∈ X and every t ∈ R. We close this section with a characterization of the infinitesimal generator of a unitary group. Theorem 6.2.5. Let A be a linear operator on a Hilbert space X. Then A is the infinitesimal generator of the unitary group (T (t))t∈R on X if and only if A and −A generate a contraction semigroup.

6.3. Exercises

75

Proof. Let (T (t))t∈R be the unitary group, then it is easy to see that

T (t)x0 = x0 ,

t > 0,

and

T (−t)x0 = x0 ,

t > 0.

Thus by Theorem 6.2.3, A and −A generate contraction semigroups. Assume next that A and −A generate a contraction semigroup, then by Theorem 6.2.3, A generates a group which is given by (6.17). Let x0 ∈ X and t > 0, then

x0 = T (t)T (−t)x0 ≤ T (−t)x0 since (T (t))t≥0 is a contraction semigroup ≤ x0 , where we have used that (T (−t))t≥0 is a contraction semigroup. From these inequalities we obtain that T (−t)x0 = x0 , t ≥ 0. Similarly, we can show that

T (t)x0 = x0 . Thus (T (t))t∈R is a unitary group. Remark 6.2.6. Another useful characterization of a unitary group is the following: A is the infinitesimal generator of a unitary group if and only if A = −A∗ , i.e., A is skew-adjoint.

6.3 Exercises 6.1. Let A0 be a self-adjoint, non-negative operator on the Hilbert space X. Prove that A := −A0 is the infinitesimal generator of a contraction semigroup on X. 6.2. Let A be the infinitesimal generator of the C0 -semigroup (T (t))t≥0 on the Hilbert space X. Prove that (T (t))t≥0 is a contraction semigroup if and only if Ax, x + x, Ax ≤ 0 for every x ∈ D(A). 6.3. Let X be the Hilbert space L2 (0, 1) and define for t ≥ 0 and f ∈ X the following operator: " f (t + ζ), ζ ∈ [0, 1], t + ζ ≤ 1, (T (t)f ) (ζ) = (6.21) 0, ζ ∈ [0, 1], t + ζ > 1. (a) Show that (T (t))t≥0 is a contraction semigroup on X. (b) Prove that the infinitesimal generator of this semigroup is given by Af =

df , dζ

(6.22)

76

Chapter 6. Contraction and Unitary Semigroups with domain + , df D(A) = f ∈ L2 (0, 1) | f is absolutely cont., ∈ L2 (0, 1) and f (1) = 0 . dζ Hint: In order the domain of the generator A, note that to calculate D(A) = ran (I − A)−1 and use Proposition 5.2.4.

6.4. Consider the following operator on X = L2 (0, 1): Af =

df , dζ

(6.23)

with domain + , df D(A) = f ∈ L2 (0, 1) | f is absolutely cont., ∈ L2 (0, 1) and f (1) = f (0) . dζ (a) Show that this operator generates a unitary group on X. (b) Find the expression for this group. 6.5. In this exercise, we give some simple p.d.e.’s which do not generate a C0 semigroup. As state space we choose again X = L2 (0, 1). (a) Let A be defined as Af =

df , dζ

(6.24)

with domain + , df ∈ L2 (0, 1) . D(A) = f ∈ L2 (0, 1) | f is absolutely cont. and dζ Show that A is not an infinitesimal generator on L2 (0, 1). Hint: Show that αI − A is not injective for any α > 0. (b) Let A be defined as Af =

df , dζ

(6.25)

with domain + , df D(A) = f ∈ L2 (0, 1) | f is absolutely cont., ∈ L2 (0, 1) and f (0) = 0 . dζ Show that A is not an infinitesimal generator on L2 (0, 1). Hint: Show that the norm of (αI − A)−1 f for f (ζ) = 1, ζ ∈ [0, 1] does not satisfy the norm bound of Proposition 5.2.4.


77

6.4 Notes and references Contraction and unitary semigroups appear often when applying semigroup theory to partial differential equations, see for instance Chapter 7. This implies that a characterization as given in the Lumer-Phillips Theorem is widely used. This theorem dates 50 years back, see [39] and [46]. There are many good references concerning this theorem, such as Curtain and Zwart [10], Engel and Nagel [15], Pazy [44], Yosida [61], and Hille and Phillips [24].

Chapter 7

Homogeneous Port-Hamiltonian Systems In the previous two chapters we have formulated partial differential equations as abstract first order differential equations. Furthermore, we described the solutions of these differential equations via a strongly continuous semigroup. These differential equations were only weakly connected to the norm of the underlying state space. However, in this chapter we consider a class of differential equations for which there is a very natural state space norm. This natural choice enables us to show that the corresponding semigroup is a contraction semigroup.

7.1 Port-Hamiltonian systems We begin by introducing a standard example of the class of port-Hamiltonian systems. Example 7.1.1. We consider the vibrating string as depicted in Figure 7.1, which we have studied in Example 1.1.4 as well. The string is fixed at the left-hand

11 00 00 11 00 11 00 11

u

Figure 7.1: The vibrating string side and may move freely at the right-hand side. We allow that a force u may be applied at the right-hand side. The model of the (undamped) vibrating string is B. Jacob and H.J. Zwart, Linear Port-Hamiltonian Systems on Infinite-dimensional Spaces, Operator Theory: Advances and Applications 223, DOI 10.1007/978-3-0348-0399-1_7, © Springer Basel 2012

79

80

Chapter 7. Homogeneous Port-Hamiltonian Systems

given by ∂2w 1 ∂ (ζ, t) = 2 ∂t ρ(ζ) ∂ζ

T (ζ)

∂w (ζ, t) , ∂ζ

(7.1)

where ζ ∈ [a, b] is the spatial variable, w(ζ, t) is the vertical position of the string at place ζ and time t, T is the Young’s modulus of the string, and ρ is the mass density, which may vary along the string. This system has the energy/Hamiltonian 1 E(t) = 2

b

ρ(ζ) a

2 2 ∂w ∂w (ζ, t) + T (ζ) (ζ, t) dζ. ∂t ∂ζ

(7.2)

Assuming that (7.1) possesses a (classical) solution, we may differentiate this Hamiltonian along the solutions of the partial differential equation

b

∂2w ∂2w ∂w ∂w (ζ, t) 2 (ζ, t) + T (ζ) (ζ, t) (ζ, t) dζ ∂t ∂t ∂ζ ∂ζ∂t a b ∂w ∂ ∂w ∂2w ∂w = (ζ, t) (ζ, t) + T (ζ) (ζ, t) (ζ, t) dζ T (ζ) ∂ζ ∂ζ ∂ζ ∂ζ∂t a ∂t ∂w ∂w ∂w ∂w = (b, t)T (b) (b, t) − (a, t)T (a) (a, t), (7.3) ∂t ∂ζ ∂t ∂ζ

dE (t) = dt

ρ(ζ)

∂w where we used integration by parts. Note that ∂w ∂t is the velocity, and T ∂ζ is the force. Furthermore, velocity times force is power which by definition also equals the change of energy. Hence, (7.3) can be considered as a power balance, and the change of internal power only happens via the boundary of the spatial domain. The string is fixed at ζ = a, that is, ∂w ∂t (a, t) = 0. Furthermore, we have assumed that we control the force at ζ = b. Thus the power balance (7.3) becomes

∂w dE (t) = (b, t)u(t). dt ∂t

(7.4)

Next we apply a transformation to equation (7.1). We define x1 = ρ ∂w ∂t (momentum) and x2 = ∂w (strain). Then (7.1) can equivalently be written as ∂ζ ∂ ∂t

x1 (ζ, t) x2 (ζ, t)

1 0 ∂ 0 1 x1 (ζ, t) ρ(ζ) = 1 0 ∂ζ x2 (ζ, t) 0 T (ζ) ∂ x1 (ζ, t) = P1 H(ζ) , x2 (ζ, t) ∂ζ

where P1 = [ 01 10 ] and H(ζ) =

1 ρ(ζ)

0

(7.5)

. The energy/Hamiltonian becomes in T (ζ) 0

7.1. Port-Hamiltonian systems the new variables, see (7.2), 1 b x1 (ζ, t)2 + T (ζ)x2 (ζ, t)2 dζ E(t) = 2 a ρ(ζ)

1 b x1 (ζ, t) x1 (ζ, t) x2 (ζ, t) H(ζ) = dζ. x2 (ζ, t) 2 a

81

(7.6)

Various examples can be written in a similar form as the previous example, which is a particular example of a port-Hamiltonian system. Definition 7.1.2. Let P1 ∈ Kn×n be invertible and self-adjoint, let P0 ∈ Kn×n be skew-adjoint, i.e., P0∗ = −P0 , and let H ∈ L∞ ([a, b]; Kn×n ) such that H(ζ)∗ = H(ζ), mI ≤ H(ζ) ≤ M I for a.e. ζ ∈ [a, b] and constants m, M > 0 independent of ζ. We equip the Hilbert space X := L2 ([a, b]; Kn ) with the inner product 1 b f, g X = g(ζ)∗ H(ζ)f (ζ) dζ. (7.7) 2 a Then the differential equation ∂x ∂ (ζ, t) = P1 (H(ζ)x(ζ, t)) + P0 (H(ζ)x(ζ, t)) . ∂t ∂ζ

(7.8)

is called a linear, first order port-Hamiltonian system. The associated Hamiltonian E : [a, b] → K is given by 1 b E(t) = x(ζ, t)∗ H(ζ)x(ζ, t)dζ. (7.9) 2 a Remark 7.1.3. We make the following observations concerning this definition. 1. Hamiltonian differential equations form an important subclass within ordinary and partial differential equations. They include linear and non-linear differential equations, and appear in many physical models. We restrict ourselves to linear differential equations of the type (7.8), and will normally omit the terms “linear, first order”. 2. In Hamiltonian differential equations, the associated Hamiltonian is normally denoted by H instead of E. Since in our examples the Hamiltonian will always be the energy, and since we want to have a clear distinction between H and the Hamiltonian, we have chosen for E. 3. Since the squared norm of X equals the energy associated to the linear, first order port-Hamiltonian system, we call X the energy space. 4. Since mI ≤ H(ζ) ≤ M I for a.e. ζ ∈ [a, b], the standard L2 -norm is equivalent to the norm defined by (7.7), i.e., for all f ∈ L2 ([a, b]; Kn ) m M

f 2L2 ≤ f 2X ≤

f 2L2 . 2 2

82


5. Real self-adjoint matrices are also called symmetric, and for complex selfadjoint the term Hermitian is used as well. We have seen that the vibrating string is a port-Hamiltonian system. Next we show that the model of the Timoshenko beam is a port-Hamiltonian system as well. Example 7.1.4. The model of the Timoshenko beam incorporates shear and rotational inertia effects in a vibrating beam. Its equations are given by ∂2w ∂ ∂w ρ(ζ) 2 (ζ, t) = (ζ, t) − φ(ζ, t) , ζ ∈ (a, b), t ≥ 0, K(ζ) ∂t ∂ζ ∂ζ (7.10) ∂w ∂2φ ∂ ∂φ Iρ (ζ) 2 (ζ, t) = (ζ, t) − φ(ζ, t) , EI(ζ) (ζ, t) + K(ζ) ∂t ∂ζ ∂ζ ∂ζ where w(ζ, t) is the transverse displacement of the beam and φ(ζ, t) is the rotation angle of a filament of the beam. The coefficients ρ(ζ), Iρ (ζ), EI(ζ), and K(ζ) are the mass per unit length, the rotary moment of inertia of a cross section, the product of Young’s modulus of elasticity and the moment of inertia of a cross section, and the shear modulus, respectively. The energy/Hamiltonian for this system is given by 2 2 ) ∂w ∂w 1 b (ζ, t) − φ(ζ, t) + ρ(ζ) (ζ, t) E(t) = K(ζ) 2 a ∂ζ ∂t 2 2 * ∂φ ∂φ +EI(ζ) (ζ, t) + Iρ (ζ) (ζ, t) dζ. (7.11) ∂ζ ∂t In order to show that the model of the Timoshenko beam is a port-Hamiltonian system, we introduce the following (physical) notation. ∂w (ζ, t) − φ(ζ, t) ∂ζ ∂w x2 (ζ, t) = ρ(ζ) (ζ, t) ∂t ∂φ x3 (ζ, t) = (ζ, t) ∂ζ ∂φ x4 (ζ, t) = Iρ (ζ) (ζ, t) ∂t x1 (ζ, t) =

shear displacement momentum angular displacement angular momentum.

Calculating the time derivative of the variables x1 , . . . , x4 , we find by using (7.10), ⎤ ⎡ x2 (ζ,t) x4 (ζ,t) ∂ ⎡ ⎤ − ∂ζ ρ(ζ) I (ζ) ρ x1 (ζ, t) ⎢ ⎥ ⎥ ⎢ ∂ ⎥ ∂ ⎢ 1 (ζ, t)) ⎢ x2 (ζ, t) ⎥ = ⎢ ∂ζ (K(ζ)x ⎥ (7.12) ⎢ ⎥ ⎦ ⎣ x (ζ,t) ∂ 4 ∂t x3 (ζ, t) ⎦ ⎣ ∂ζ Iρ (ζ) x4 (ζ, t) ∂ ∂ζ (EI(ζ)x3 (ζ, t)) + K(ζ)x1 (ζ, t)


83

Dropping for readability the coordinates ζ and t, we obtain ⎡

⎤ ⎡ 0 x1 ⎥ ⎢ ∂ ⎢ ⎢ x2 ⎥ = ⎢ 1 ∂t ⎣ x3 ⎦ ⎣ 0 x4 0 ⎡

1 0 0 0

0 ⎢ 0 +⎢ ⎣ 0 1

0 0 0 1 0 0 0 0

⎤ ⎛⎡ K 0 0 ⎥ ⎢ ⎜ 0 ρ1 ∂ 0 ⎥ ⎜⎢ 1 ⎦ ∂ζ ⎝⎣ 0 0 0 0 0 ⎤⎡ K 0 0 −1 ⎢ 0 ρ1 0 0 ⎥ ⎥⎢ 0 0 ⎦⎣ 0 0 0 0 0 0

0 0 EI 0 0 0 EI 0

⎤⎡ ⎤⎞ 0 x1 0 ⎥ ⎢ x2 ⎥⎟ ⎥⎢ ⎥⎟ 0 ⎦ ⎣ x3 ⎦⎠ 1 x4 Iρ ⎤⎡ ⎤ 0 x1 0 ⎥ ⎢ x2 ⎥ ⎥⎢ ⎥. 0 ⎦ ⎣ x3 ⎦ 1 x4 Iρ

(7.13)

Formulating the energy/Hamiltonian in the variables x1 , . . . , x4 is easier, see (7.11) E(t) =

1 2

1 = 2

b

1 1 x2 (ζ, t)2 + EI(ζ)x3 (ζ, t)2 + x4 (ζ, t)2 dζ ρ(ζ) Iρ (ζ) ⎤⎡ ⎤∗⎡ ⎤⎞ K(ζ) 0 0 0 x1 (ζ, t) 1 ⎥⎢ 0 0 0 ⎥⎢ x2 (ζ, t) ⎥⎟ ρ(ζ) ⎥⎢ ⎥⎢ ⎥⎟ dζ. ⎦⎣ 0 0 EI(ζ) 0 ⎦⎣ x3 (ζ, t) ⎦⎠ 1 0 0 0 x4 (ζ, t) Iρ (ζ) (7.14)

K(ζ)x1 (ζ, t)2 + a

a

⎛⎡ b

x1 (ζ, t) ⎜⎢ x2 (ζ, t) ⎜⎢ ⎝⎣ x3 (ζ, t) x4 (ζ, t)

This shows that the Timoshenko beam can be written as a port-Hamiltonian system on X = L2 ([a, b]; R4 ). Next we calculate the power. Using the above notation and the model (7.10), we find that the power equals b dE x2 (ζ, t) x4 (ζ, t) (t) = K(ζ)x1 (ζ, t) + EI(ζ)x3 (ζ, t) . dt ρ(ζ) Iρ (ζ) a

(7.15)

Again, the power goes via the boundary of the spatial domain. As we have seen in the examples, the change of energy (power) of these systems was only possible via the boundary of its spatial domain. In the following theorem we show that this is a general property for any system which is of the form (7.8) with Hamiltonian (7.9). Theorem 7.1.5. Let x be a classical solution of the port-Hamiltonian system (7.8) with Hamiltonian (7.9). Then the following balance equation holds:

b 1 dE ∗ (t) = (H(ζ)x(ζ, t)) P1 H(ζ)x(ζ, t) a . dt 2

(7.16)

84


Proof. By using the partial differential equation, we obtain ∂x 1 b ∂x (ζ, t)∗ H(ζ)x(ζ, t)dζ + x(ζ, t)∗ H(ζ) (ζ, t)dζ 2 a ∂t a ∂t ∗ b 1 ∂ = P1 (H(ζ)x(ζ, t)) + P0 H(ζ)x(ζ, t) H(ζ)x(ζ, t)dζ 2 a ∂ζ 1 b ∂ ∗ + x(ζ, t) H(ζ) P1 (H(ζ)x(ζ, t)) + P0 H(ζ)x(ζ, t) dζ. 2 a ∂ζ

dE 1 (t) = dt 2

b

Using now the fact that P1 and H(ζ) are self-adjoint, and P0 is skew-adjoint, we write the last expression as 1 2

a

1 + 2

b

∗ ∂ ∂ ∗ (H(ζ)x(ζ, t)) P1 H(ζ)x(ζ, t) + (H(ζ)x(ζ, t)) P1 (H(ζ)x(ζ, t)) dζ ∂ζ ∂ζ

b

∗

∗

− (H(ζ)x(ζ, t)) P0 H(ζ)x(ζ, t) + (H(ζ)x(ζ, t)) P0 H(ζ)x(ζ, t)dζ a

1 b ∂ ∗ = (H(ζ)x(ζ, t)) P1 H(ζ)x(ζ, t) dζ 2 a ∂ζ

b 1 ∗ (H(ζ)x(ζ, t)) P1 H(ζ)x(ζ, t) a . = 2

Hence we have proved the theorem.

The balance equation (7.16) is very important, and guides us in many problems. It explains the name port-Hamiltonian system. The system has a Hamiltonian (most times energy) and changes of this quantity can only occur via the boundary, i.e., the ports to the outside world. Note that (7.8) with P0 = 0 is the infinitedimensional counterpart of the finite-dimensional port-Hamiltonian system of Sec∂ tion 2.3. The J is replaced by P1 ∂ζ , which is a skew-symmetric operator, and H is replaced by the operator, which multiplies by H(·). As we showed in the previous two chapters, a partial differential equation needs boundary conditions in order to possess a unique solution. In the next section we characterize those boundary conditions for which the port-Hamiltonian system generates a contraction semigroup.

7.2 Generation of contraction semigroups In this section, we apply the general results presented in Chapters 5 and 6 to port-Hamiltonian systems, i.e., we consider partial differential equations of the form ∂x ∂ (ζ, t) = P1 (H(ζ)x(ζ, t)) + P0 H(ζ)x(ζ, t). (7.17) ∂t ∂ζ

7.2. Generation of contraction semigroups

85

and we aim to characterize (homogeneous) boundary conditions such that (7.17) possesses a unique solution. In order to write (7.17) as an abstract differential equation, we “hide” the spatial dependence, and we write the p.d.e. as the (abstract) ordinary differential equation dx ∂ (t) = P1 (Hx(t)) + P0 (Hx(t)) . dt ∂ζ

(7.18)

Hence we consider the operator A0 x := P1

d (Hx) + P0 (Hx) dζ

(7.19)

on the state space X = L2 ([a, b]; Kn ) with inner product f, g X = and domain

1 2

b

(7.20)

g(ζ)∗ H(ζ)f (ζ) dζ,

(7.21)

a

3 4 D(A0 ) = x ∈ X | Hx ∈ H 1 ([a, b]; Kn ) .

(7.22)

Here H 1 ([a, b]; Kn ) is the vector space of all functions from [a, b] to Kn , which are square integrable, absolutely continuous, and the derivative is again square integrable, that is, H 1 ([a, b]; Kn ) = {f ∈ L2 ([a, b]; Kn ) | f is absolutely continuous and

df ∈ L2 ([a, b]; Kn )}. dζ

Note, that A0 as the maximal domain. In order to guarantee that (7.17) possesses a unique solution we have to add boundary conditions. It turns out that it is better to formulate them in the boundary effort and boundary flow, which are defined as 1 e∂ = √ ((Hx)(b) + (Hx)(a)) 2

1 and f∂ = √ (P1 (Hx)(b) − P1 (Hx)(a)) , 2 (7.23) respectively. Next, we show some properties of the operator A0 . Lemma 7.2.1. Consider the operator A0 defined in (7.19) and (7.22) associated to a port-Hamiltonian system, that is, the assumptions of Definition 7.1.2 are satisfied. Then the following results hold: 1. Re A0 x, x X =

1 ∗ (f e∂ + e∗∂ f∂ ). 4 ∂

2. For every [ uy ] ∈ K2n there exists an x0 ∈ D(A0 ) such that

(Hx0 )(b) (Hx0 )(a)

= [ uy ].

86


Proof. 1. For the differential operator A0 and x0 ∈ D(A0 ) we have A0 x, x X + x, A0 x X

∂ x(ζ) H(ζ) P1 (Hx)(ζ) + P0 (Hx)(ζ) dζ ∂ζ a ∗ b 1 ∂ + P1 (Hx)(ζ) + P0 (Hx)(ζ) H(ζ)x(ζ) dζ. 2 a ∂ζ

1 = 2

b

∗

Using the fact that P1 is self-adjoint, and P0 is skew-adjoint, we find A0 x, x X + x, A0 x X ∗ d 1 b d ∗ (Hx) (ζ) P1 H(ζ)x(ζ)dζ = (H(ζ)x(ζ)) P1 (Hx) (ζ) + 2 a dζ dζ 1 b ∗ ∗ + (H(ζ)x(ζ)) (P0 H(ζ)x(ζ)) − (H(ζ)x(ζ)) P0 H(ζ)x(ζ)dζ 2 a 1 b d ∗ = (Hx) (ζ)P1 (Hx) (ζ) dζ 2 a dζ 1 ∗ ∗ = (Hx) (b)P1 (Hx) (b) − (Hx) (a)P1 (Hx) (a) . 2 Combining this equality with (7.23), we obtain A0 x, x X + x, A0 x X =

1 ∗ (f e∂ + e∗∂ f∂ ) . 2 ∂

(7.24)

2. It is easy to see that x0 defined as x0 (ζ) = H

−1

ζ −a (ζ) y + u b−a

is an element of the domain of A0 and satisfies the boundary conditions.

The balance equation (7.16) shows that the boundary flow is not determined by x, but by Hx. Therefore, we formulate the boundary conditions in this variable. So we consider the boundary conditions ˜ B H(b)x(b, t) = 0, W t ≥ 0. (7.25) H(a)x(a, t) It turns out that formulating the boundary conditions directly in x or Hx at ζ = a and ζ = b is not the best choice for characterizing generators of contraction semigroups. It is better to formulate them in the boundary effort and boundary flow, which are defined by (7.23). We write this as a matrix vector product, i.e., (Hx)(b) f∂ = R0 , (7.26) e∂ (Hx)(a)


87

with R0 ∈ K2n×2n defined as −P1 . I

1 P1 R0 = √ 2 I

(7.27)

Next, we show some properties of this transformation. Lemma 7.2.2. Let P1 be self-adjoint and invertible in Kn×n , then the matrix R0 ∈ K2n×2n defined by (7.27) is invertible, and satisfies P1 0 = R0∗ ΣR0 , (7.28) 0 −P1 where

Σ=

0 I

I 0

.

(7.29)

All possible matrices R which satisfy (7.28) are given by the formula R = U R0 , with U satisfying U ∗ ΣU = Σ. Proof. We have that 1 P1 I 0 √ I 2 −P1 I

I 0

P1 I

−P1 I

1 √ = 2

P1 0

0 −P1

.

1 Thus using the fact that P1 is self-adjoint, we obtain that R0 := √12 PI1 −P I satisfies (7.28). Since P1 is invertible, the invertibility of R0 follows from equation (7.28). Let R be another solution of (7.28). Hence P1 0 R∗ ΣR = = R0∗ ΣR0 . 0 −P1 This can be written in the equivalent form R0−∗ R∗ ΣRR0−1 = Σ. Calling RR0−1 = U , we obtain U ∗ ΣU = Σ and R = U R0 , which proves the assertion. Since the matrix R0 is invertible, we can write any condition which is formulated in (Hx)(b) and (Hx)(a) into an equivalent condition which is formulated in f∂ and e∂ . Using (7.26), we write the boundary condition (7.25) (equivalently) as f∂ (t) WB = 0, (7.30) e∂ (t)

88


˜ B R−1 . Thus we study the operator where WB = W 0 Ax := P1

d (Hx) + P0 (Hx) dζ

(7.31)

with domain D(A) = {x ∈ L2 ([a, b]; Kn ) | Hx ∈ H 1 ([a, b]; Kn ), WB

f

∂

e∂

= 0}.

(7.32)

In Theorem 7.2.4 we characterize the boundary conditions for which the operator (7.31) with domain (7.32) generates a contraction semigroup. The following technical lemma is useful. Lemma 7.2.3. Let Z be a Hilbert space with inner product ·, · , and let P ∈ L(Z) be a coercive operator on Z, i.e., P is self-adjoint and P > εI, for some ε > 0. We define ZP as the Hilbert space Z with the inner product ·, · P := ·, P· . Then the operator A with domain D(A) generates a contraction semigroup on Z if and only if AP with domain D(AP) = {z ∈ Z | Pz ∈ D(A)} generates a contraction semigroup on ZP . Proof. By the coercivity of P it is easy to see that ZP is a Hilbert space. Since the inverse of P is also coercive, we only have to prove one implication. We assume that A with domain D(A) generates a contraction semigroup. We first show that Rex, APx P ≤ 0 for x ∈ D(AP). For x ∈ D(AP), we have that Rex, APx P = Rex, PAPx = RePx, APx ≤ 0, where we have used that P is self-adjoint and A is dissipative. Clearly, AP is a linear, densely defined and closed operator on ZP . Using Theorem 6.1.8 it remains to show that the adjoint of AP is dissipative as well. z ∈ ZP is in the domain of the adjoint of AP if and only if there exists a w ∈ ZP such that z, APx P = w, x P

(7.33)

for all x ∈ D(AP). Equation (7.33) is equivalent to Pz, APx = w, Px .

(7.34)

x ∈ D(AP) if and only if Px ∈ D(A), which implies that (7.34) holds for all x ∈ D(AP) if and only if Pz, A˜ x = w, x˜ (7.35) holds for all x˜ ∈ D(A). Equation (7.35) is equivalent to the fact that Pz ∈ D(A∗ ) and w = A∗ Pz. Summarizing, z ∈ D((AP)∗ ) if and only if Pz ∈ D(A∗ ) and furthermore, ∗

(AP) z = A∗ Pz,

z ∈ D ((AP)∗ ) .

(7.36)


89

It remains to show that the adjoint as defined in (7.36) is dissipative. This is straightforward by using (7.36) Rex, (AP)∗ x P = Rex, PA∗ Px = RePx, A∗ Px ≤ 0, where we used the fact that A∗ is dissipative.

The following theorem characterizes the matrices WB for which the operator A with domain (7.32) generates a contraction semigroup. The matrix Σ is defined in (7.29). For the proof of this theorem we need some results concerning matrices, which can be found in Section 7.3. Theorem 7.2.4. Consider the operator A defined in (7.31) and (7.32) associated to a port-Hamiltonian system, that is, the assumptions of Definition 7.1.2 are ˜ B , is an n × 2n matrix of rank n. satisfied. Furthermore, WB , or equivalently W Then the following statements are equivalent. 1. A is the infinitesimal generator of a contraction semigroup on X. 2. Re Ax, x X ≤ 0 for every x ∈ D(A). 3. WB ΣWB∗ ≥ 0. Proof. The Lumer-Phillips Theorem shows that part 1 implies part 2. We next show that part 2 implies part 3. Thus we assume that Re Ax, x X ≤ 0 for every x ∈ D(A). Using the fact that A is a restriction of A0 , Lemma 7.2.1 implies f∂∗ e∂ + e∗∂ f∂ ≤ 0 for every x ∈ D(A). Furthermore, by Lemma 7.2.1 for every pair [ fe ] ∈ ker WB , there exists a function x ∈ D(A) with boundary effort e∂ = e and boundary flow f∂ = f . Thus we have f ∗ e + e∗ f ≤ 0 for every pair [ fe ] ∈ ker WB .

We write WB as WB = W1 W2 . If y lies in the kernel of W1 + W2 , then WB [ yy ] = 0, and thus y ∗ y + y ∗ y ≤ 0, which implies y = 0. This shows that the matrix W1 + W2 is injective, and hence invertible. Defining V := (W1 + W2 )−1 (W1 − W2 ), we have

W1

W2

=

1 (W1 + W2 ) I + V 2

I −V

.

Let [ fe ] ∈ ker WB be arbitrary. By Lemma 7.3.2 there exists a vector such that I−V . This implies [ fe ] = −I−V 0 ≥ f ∗ e + e∗ f = ∗ (−2I + 2V ∗ V ) ,

(7.37)

This inequality holds for any [ fe ] ∈ ker WB . Since the n×2n matrix WB has rank

n, I−V for its kernel has dimension n, and so the set of vectors satisfying [ fe ] = −I−V some [ fe ] ∈ ker WB equals the whole space Kn . Thus (7.37) implies that V ∗ V ≤ I, and by Lemma 7.3.1 we obtain WB ΣWB∗ ≥ 0.

90


Finally, we show that part 3 implies part 1. The proof of this implication is divided in several steps. Using Lemma 7.3.1, we write the matrix WB as WB = S I + V I − V , where S is invertible and V ∗ V ≤ I. Step 1. Using (7.31) and (7.32) we write Ax = AI Hx, where AI is defined by (7.31) and (7.32) with H ≡ I. Furthermore, the inner product of the state space X, see (7.7), equals f, Hg I , where ·, · I is the inner product (7.7) with H ≡ I, i.e., the standard inner product on L2 ([a, b]; Kn ). Since H is a coercive multiplication operator on L2 ([a, b]; Kn ), we may apply Lemma 7.2.3 to conclude that A generates a contraction semigroup on X if and only if AI generates a contraction semigroup on L2 ([a, b]; Kn ). Hence, it is sufficient to prove this implication for H = I. Step 2. Let H = I and x ∈ D(A). Lemma 7.2.1 implies Ax, x X + x, Ax X =

1 ∗ (f e∂ + e∗∂ f∂ ) . 2 ∂

By assumption, the vector fe∂∂ lies in the kernel of WB . Using Lemma 7.3.2, fe∂∂ I−V

equals −I−V for some ∈ Kn . Thus we get 1 ∗ (f e∂ + e∗∂ f∂ ) 2 ∂ 1 = ( ∗ (I − V ∗ )(−I − V ) + ∗ (−I − V ∗ )(I − V ) ) 2 = ∗ (−I + V ∗ V ) ≤ 0, (7.38)

Ax, x X + x, Ax X =

where we used again Lemma 7.3.1. This implies the dissipativity of A. Step 3. By the Lumer-Phillips Theorem (Theorem 6.1.7) it remains to show that the range of I − A equals X. The equation (I − A)x = y is equivalent to the differential equation x(ζ) − P1 x(ζ) ˙ − P0 x(ζ) = y(ζ),

ζ ∈ [a, b].

(7.39)

Thanks to the invertibility of the matrix P1 , the solution of (7.39) is given by ζ −1 −1 (P1−1 −P1−1 P0 )(ζ−a) x(a) − e(P1 −P1 P0 )(ζ−τ ) P1−1 y(τ )dτ. (7.40) x(ζ) = e a

x ∈ D(A) if and only if, see (7.26) and (7.30), x(b) Ex(a) + q 0 = WB R0 = WB R0 , x(a) x(a) −1

−1

−1

−1

(7.41)

where E = e(P1 −P1 P0 )(b−a) and q = − a e(P1 −P1 P0 )(b−τ ) P1−1 y(τ )dτ , see (7.40). Equation (7.41) can be equivalently written as E q x(a) = −WB R0 . (7.42) WB R0 I 0 b


91

Next we prove that the square matrix WB R0 [ EI ] is invertible. A square matrix is invertible if and only if it is injective. Therefore, we assume that there exists a vector r0 = 0 such that E WB R0 r0 = 0. (7.43) I Now we solve equation (7.39) for x(a) = r0 and y ≡ 0. The corresponding solution lies in D(A) by (7.40) and (7.41). Thus (I − A)x = 0 has a non-zero solution. In other words, x, being the solution of (7.39) with x(a) = r0 and y ≡ 0, is an eigenfunction of A with eigenvalue one. However, by (7.38), A possesses no eigenvalues in the right half-plane, which leads to a contradiction, implying that WB R0 [ EI ] is invertible, and so (7.41) has a unique solution in the domain of A. The function y was arbitrary, and so we have proved that the range of I − A equals X. Using the Lumer-Phillips Theorem 6.1.7, we conclude that A generates a contraction semigroup. We apply this theorem to our vibrating string example, see Example 7.1.1. Example 7.2.5. Consider the vibrating string on the spatial domain [a, b], see Example 7.1.1. The string is fixed at the left-hand side and we apply no force at the right-hand side, which gives the boundary conditions ∂w (a, t) = 0, ∂t

∂w (b, t) = 0. (7.44) ∂ζ 1 0 We have P1 = [ 01 10 ], P0 = 0, and H(ζ) = ρ(ζ) , see (7.5). Using this and 0 T (ζ) T (b)

equation (7.23) the boundary variables are given by ∂w ∂w ∂w 1 1 T (b) ∂w ∂ζ (b) − T (a) ∂ζ (a) ∂t (b) + ∂t (a) √ f∂ = √ , e = . ∂ ∂w ∂w ∂w ∂w 2 2 T (b) ∂ζ (b) + T (a) ∂ζ (a) ∂t (b) − ∂t (a) (7.45) The boundary condition (7.44) becomes in these variables T (b) ∂w 0 ∂ζ (b, t) = ∂w 0 (a, t) ∂t 1 1 0 0 1 f∂ (t) f∂ (t) = √ = WB , (7.46) e∂ (t) e∂ (t) 2 0 −1 1 0 0 0 1

with WB = √12 10 −1 1 0 . Since WB is a 2 × 4 matrix with rank 2, and since T WB ΣWB = 0, we conclude from Theorem 7.2.4 that the operator associated to the p.d.e. generates a contraction semigroup on L2 ([a, b]; R2 ) with the norm (7.6). In the above we used part 3 of Theorem 7.2.4 to show that the operator associated to the p.d.e. (7.1) with boundary conditions (7.44) generates a contraction semigroup on the energy space. To complete this example we show how this can be proved using part 2 of Theorem 7.2.4.

92

Chapter 7. Homogeneous Port-Hamiltonian Systems From (7.25) and (7.44) we see that T (b) ∂w (b, t) 0 0 1 0 ∂ζ = = ∂w 0 0 0 1 (a, t) ∂t

0 0

(Hx)(b) (Hx)(b)

˜B =W

(Hx)(b) . (Hx)(b) (7.47)

˜ B is two. The operator A is given as It is clear that the rank of W 5 d 6 x2 ) d dζ (T Ax = P1 (Hx) = x1 d dζ dζ ρ

(7.48)

with domain D(A) = {x ∈ L2 ([a, b]; R2 ) | Hx ∈ H 1 ([a, b]; R2 ), and (7.47) holds}. Thus for x ∈ D(A) we have d x1 d x1 (T x2 )(ζ) · (ζ) + (ζ) · (T x2 )(ζ)dζ ρ dζ ρ a dζ b x1 = (T x2 )(ζ) (ζ) = 0 − 0 = 0, ρ a

b

Ax, x =

where we used the boundary conditions (7.47). By part 2 of Theorem 7.2.4, we conclude that A generates a contraction semigroup on X.

7.3 Technical lemmas This section contains two technical lemmas on matrix representations. They are important for the proof of Theorem 7.2.4, but not for the understanding of the examples. Lemma 7.3.1. Let W be a n × 2n matrix and let Σ = [ I0 I0 ]. Then W has rank n and W ΣW ∗ ≥ 0 if and only if there exist a matrix V ∈ Kn×n and an invertible matrix S ∈ Kn×n such that

W =S I +V I −V (7.49) with V V ∗ ≤ I, or equivalently V ∗ V ≤ I. Proof. Sufficiency: If W is of the form (7.49), then we find

I +V∗ W ΣW ∗ = S I + V I − V Σ S ∗ = S(2I − 2V V ∗ )S ∗ , I −V∗

(7.50)

which is non-negative, since V V ∗ ≤ I. Assuming rk W < n, there exists x = 0 such that x∗ W = 0, or equivalently (S ∗ x)∗ I + V I − V = 0. This implies

7.4. Exercises

93

for x ˜ := S ∗ x that x˜(I + V ) = 0 and x˜(I − V ) = 0, and thus x˜ = 0, which leads to a contradiction. Therefore, rk W = n.

Necessity: Writing W as W = W1 W2 , we see that W ΣW ∗ ≥ 0 is equivalent to W1 W2∗ + W2 W1∗ ≥ 0. Hence (W1 + W2 )(W1 + W2 )∗ ≥ (W1 − W2 )(W1 − W2 )∗ ≥ 0.

(7.51)

If x ∈ ker((W1 +W2 )∗ ), then the above inequality implies that x ∈ ker((W1 −W2 )∗ ). Thus x ∈ ker(W1∗ )∩ker(W2∗ ). Since W has full rank, this implies that x = 0. Hence W1 + W2 is invertible. Using (7.51) once more, we see that (W1 + W2 )−1 (W1 − W2 )(W1 − W2 )∗ (W1 + W2 )−∗ ≤ I and thus V := (W1 + W2 )−1 (W1 − W2 ) satisfies V V ∗ ≤ I. Summarizing, we have

W1

W2

1 W1 + W2 + W1 − W2 W1 + W2 − W1 + W2 2

1 = (W1 + W2 ) I + V I − V . 2 =

Defining S := 12 (W1 + W2 ), we have shown the representation (7.49).

Lemma 7.3.2. Suppose that the n × 2n matrix W can be written in the format of equation (7.49), i.e., W = S[I + V I − V ] with S and V square matrices, and S I−V

. invertible. Then the kernel of W equals the range of −I−V I−V

Proof. Let [ xx12 ] be in the range of −I−V . By the equality (7.49), we have that

x1 x1 W =S I +V I −V x2 x2

I −V =S I +V I −V = 0. −I − V I−V

lies in the kernel of W . The conditions on W imply Hence the range of −I−V that rk W = n, and so the kernel

of W has dimension n. It is sufficient to show I−V that the 2n × n-matrix −I−V has full rank. If this is not the case, then there is a non-trivial element in its kernel. It is easy to see that the kernel consists of zero only, and thus we have proved the lemma.

7.4 Exercises 7.1. Consider the transmission line on the spatial interval [a, b]: ∂ ∂Q (ζ, t) = − ∂t ∂ζ ∂φ ∂ (ζ, t) = − ∂t ∂ζ

φ(ζ, t) , L(ζ) Q(ζ, t) . C(ζ)

(7.52)

94

Chapter 7. Homogeneous Port-Hamiltonian Systems Here Q(ζ, t) is the charge at position ζ ∈ [a, b] and time t > 0, and φ(ζ, t) is the (magnetic) flux at position ζ and time t. C is the (distributed) capacity and L is the (distributed) inductance. V (a)

V (b)

I(a) a

b

I(b)

Figure 7.2: Transmission line The voltage and current are given by V = Q/C and I = φ/L, respectively. The energy of this system is given by 1 b φ(ζ, t)2 Q(ζ, t)2 E(t) = + dζ. (7.53) 2 a L(ζ) C(ζ) Formulate the transmission line as depicted in Figure 7.2 as a port-Hamiltonian system, see Definition 7.1.2. 7.2. Consider the operator A defined by (7.31) and (7.32) associated to a portHamiltonian system, that is, the assumptions of Definition 7.1.2 are satisfied. Furthermore, assume that WB is a full rank n × 2n matrix. Show that the following are equivalent: (a) A is the infinitesimal generator of a unitary group on X. (b) ReAx, x X = 0 for all x ∈ D(A). (c) WB ΣWB∗ = 0. 7.3. Let L2 ([a, b]; Kn ) be the standard Lebesgue space with its standard inner product. Consider the operator A defined by (7.31) and (7.32) with WB a full rank n × 2n matrix satisfying WB ΣWB∗ ≥ 0. Further we assume that the assumptions of Definition 7.1.2 are satisfied. Show that A generates a strongly continuous semigroup on L2 ([a, b]; Kn ). Is this semigroup a contraction semigroup? 7.4. Consider coupled vibrating strings as given in the figure below. We assume that the length of all strings are equal. The model for every vibrating string is given by (1.13) with physical parameters, ρI , TI , ρII , etc. Furthermore, we assume that the three strings are connected via a (mass-less) bar, as shown in Figure 7.3. This bar can only move in the vertical direction. This implies that the velocity of string I at its right-hand side equals those of the other two strings at their left-hand side. Furthermore, the force of string I at its

7.4. Exercises

11 00 00 11 00 11 00 11

I

1 0 0 1 0 1 0 1

II

III

Figure 7.3: Coupled vibrating strings

11 00 00 11 0 1 00 11 0 1 00 11 0 1 00 11 0 1 00 11 0 1 00 11 0 1 00 11 0 1 0 1

95

right-end side equals the sum of the forces of the other two at their left-hand side, i.e., ∂wI ∂wII ∂wIII TI (b) (b) = TII (a) (a) + TIII (a) (a). ∂ζ ∂ζ ∂ζ As depicted, the strings are attached to a wall. (a) Identify the boundary conditions for the system given in Figure 7.3. (b) Formulate the coupled strings as a port-Hamiltonian system (7.17) and (7.25). Furthermore, determine the energy space X. (c) Show that the differential operator associated to the above system generates a contraction semigroup on the energy space X. 7.5. In this exercise we show that a second order differential equation associated to a Sturm-Liouville operator determines a port-Hamiltonian system. The Sturm-Liouville operator is the differential operator ASL on L2 (a, b) given by 1 d dh ASL h = − (p ) + qh , w dζ dζ where w ∈ C([a, b]), p ∈ C 1 ([a, b]) and q ∈ C([a, b]) are real-valued functions with p(ζ) > 0 and w(ζ) > 0 for every ζ ∈ [a, b]. We define the domain of ASL as follows: + dh D(ASL ) = are absolutely continuous, h ∈ L2 (a, b) | h, dζ d2 h dh ∈ L2 (a, b), β1 h(a) + γ1 (a) = 0, and 2 dζ dx , dh β2 h(b) + γ2 (b) = 0 , dx where we suppose that β1 , β2 , γ1 , and γ2 are real constants satisfying |β1 | + |γ1 | > 0, and |β2 | + |γ2 | > 0. Assume now that q is identically zero.

96

Chapter 7. Homogeneous Port-Hamiltonian Systems (a) Write the second order differential equation ∂2h = −ASL h ∂t2 as a port-Hamiltonian system. What is the energy space? Hint: See Example 7.1.1. (b) Determine necessary and sufficient conditions on the constants β1 , β2 , γ1 , and γ2 such that ASL generates a contraction semigroup on the energy space.

7.5 Notes and references In this chapter we closely follow the article [36] by Le Gorrec, Zwart, and Maschke, although we present a new proof of Theorem 7.2.4. Strangely enough the class of linear, first order port-Hamiltonian systems has not been studied before within the infinite-dimensional system theory community. This class was introduced using the language of differential forms in van der Schaft and Maschke [57]. Although the class of linear, first order port-Hamiltonian systems was not studied within systems theory, one may find results on a strongly related class of operators in the literature, for instance in [19]. In [19], the authors are mainly interested in self-adjoint extensions of symmetric operators, that is, the question is to determine those boundary conditions for which an operator is self-adjoint. Note that the multiplication of a self-adjoint operator by i is a skew-adjoint operator. Skewadjoint operators are the infinitesimal generators of unitary semigroups, see Remark 6.2.6. In order to apply results from [19] we have to multiply the operators by i. The result formulated in Lemma 7.2.3 for Banach spaces can be traced back to Gustafson and Lumer [20]. However, we present a more direct proof for the Hilbert space situation.

Chapter 8

Stability This chapter is devoted to stability of abstract differential equations as well as to spectral projections and invariant subspaces. One of the most important aspects of systems theory is stability, which is closely connected to the design of feedback controls. For infinite-dimensional systems there are different notions of stability such as strong stability, polynomial stability, and exponential stability. In this chapter, we restrict ourselves to exponential stability. Strong stability will be defined in an exercise, where it is also shown that strong stability is weaker than exponential stability. The concept of invariant subspaces, which we discuss in the second part of this chapter will play a key role in the study of stabilizability in Chapter 10.

8.1 Exponential stability We start with the definition of exponential stability. Definition 8.1.1. The C0 -semigroup (T (t))t≥0 on the Hilbert space X is exponentially stable if there exist positive constants M and α such that

T (t) ≤ M e−αt

for t ≥ 0.

(8.1)

The constant α is called the decay rate, and the supremum over all possible values of α is the stability margin of (T (t))t≥0 ; this is minus its growth bound (see Theorem 5.1.5). If (T (t))t≥0 is exponentially stable, then the solution to the abstract Cauchy problem x(t) ˙ = Ax(t), t ≥ 0, x(0) = x0 , (8.2) tends to zero exponentially fast as t → ∞. Datko’s lemma is an important criterion for exponential stability. B. Jacob and H.J. Zwart, Linear Port-Hamiltonian Systems on Infinite-dimensional Spaces, Operator Theory: Advances and Applications 223, DOI 10.1007/978-3-0348-0399-1_8, © Springer Basel 2012

97

98

Chapter 8. Stability

Lemma 8.1.2 (Datko’s lemma). The C0 -semigroup (T (t))t≥0 on the Hilbert space X is exponentially stable if and only if for every x ∈ X there holds ∞

T (t)x 2 dt < ∞. (8.3) 0

Proof. The necessity is obvious, so suppose that (8.3) holds. Now Theorem 5.1.5 implies that there exist numbers M > 0 and ω > 0 such that

T (t) ≤ M eωt

for t ≥ 0.

(8.4)

Thus for every n ∈ N the operator Qn , defined by (Qn x)(t) := ½[0,n] (t)T (t)x, is a bounded linear operator from X to L2 ([0, ∞); X). (8.3) implies that for every x ∈ X the family {Qn x, n ∈ N} is uniformly bounded in n, and thus by the Uniform Boundedness Theorem, it follows that

Qn ≤ γ

(8.5)

for some γ independent of n. For 0 ≤ t ≤ 1, we have that T (t) ≤ M eωt ≤ M eω . For t > 1, we calculate t 1 − e−2ωt

T (t)x 2 = e−2ωs T (t)x 2 ds 2ω 0 t ≤ e−2ωs T (s) 2 T (t − s)x 2 ds 0 t ≤ M2

T (t − s)x 2 ds using (8.4) 0 t = M2

T (s)x 2 ds ≤ M 2 γ 2 x 2 , 0

where we used (8.5). Thus for some K > 0 and all t ≥ 0, we obtain

T (t) ≤ K and, moreover,

t

T (t)x 2 ds

t T (t)x 2 = 0

≤

t

T (s) 2 T (t − s)x 2 ds ≤ K 2 γ 2 x 2 0

Hence

Kγ

T (t) ≤ √ , t

using (8.5).

8.1. Exponential stability

99

which implies that T (τ ) < 1 for a sufficiently large τ . Consequently, log( T (τ ) ) ˜ , α > 0 such that < 0, and thus by Theorem 5.1.5 there exist M ˜ e−αt

T (t) ≤ M

for all t ≥ 0.

Lemma 8.1.2 can be used to prove a Lyapunov-type result, which will be of use in establishing stability of the abstract differential equation. Theorem 8.1.3. Suppose that A is the infinitesimal generator of the C0 -semigroup (T (t))t≥0 on the Hilbert space X. Then the following are equivalent 1. (T (t))t≥0 is exponentially stable; 2. There exists a positive operator P ∈ L(X) such that Ax, P x + P x, Ax = −x, x

for all x ∈ D(A).

(8.6)

3. There exists a positive operator P ∈ L(X) such that Ax, P x + P x, Ax ≤ −x, x

for all x ∈ D(A).

(8.7)

Equation (8.6) is called a Lyapunov equation. Proof. 1 ⇒ 2: Since (T (t))t≥0 is exponentially stable, the following operator is well defined ∞ x1 , P x2 = T (t)x1 , T (t)x2 dt. (8.8) 0

Since

|x1 , P x2 | ≤

∞

T (t)x1

T (t)x2 dt ≤

0

=

∞

M 2 e−2αt x1

x2 dt

0

M2

x1

x2 , 2α

P is bounded. Furthermore, x, P x =

∞

T (s)x 2 ds ≥ 0

0

and x, P x = 0 implies that T (t)x = 0 on [0, ∞) almost everywhere. The strong continuity of (T (t))t≥0 implies that x = 0. Thus P > 0. It remains to show that P satisfies (8.6). Using (8.8), we have for x ∈ D(A) ∞ Ax, P x + P x, Ax = T (t)Ax, T (t)x + T (t)x, T (t)Ax dt 0 ∞ dT (t)x, T (t)x = dt = 0 − x, x , dt 0

100


where we have used the exponential stability of (T (t))t≥0 . Thus P defined by (8.8) is a solution of (8.6). 2 ⇒ 3 is trivial, and thus it remains to prove 3 ⇒ 1: Suppose that there exists a bounded P > 0 such that (8.7) is satisfied. We introduce the following Lyapunov functional: V (t, x) = P T (t)x, T (t)x . Since P is positive, V (t, x) ≥ 0 for all t ≥ 0. For x ∈ D(A), we may differentiate V to obtain by (8.7) dV (t, x) = P AT (t)x, T (t)x + P T (t)x, AT (t)x ≤ − T (t)x 2 . dt Integration yields

t

0 ≤ V (t, x) ≤ V (0, x) −

T (s)x 2 ds 0

and hence t

T (s)x 2 ds ≤ V (0, x) = P x, x

for all t ≥ 0 and x ∈ D(A).

0

This inequality can be extended to all x ∈ X, since D(A) is dense in X. In other words, for every x ∈ X we have that ∞

T (s)x 2 ds ≤ P x, x < ∞ 0

and Datko’s Lemma 8.1.2 completes the proof.

In finite dimensions, one usually examines exponential stability via the spectrum of the operator. However, this is not feasible in infinite dimensions. While the inequality log T (t)

= ω0 sup Re(λ) ≤ lim (8.9) t→∞ t λ∈σ(A) always holds (Proposition 5.2.4), one need not necessarily have equality. For an example we refer the reader to [62] or [10, Example 5.1.4]. Although the location of the spectrum is not sufficient to determine the stability, the uniform boundedness of the resolvent operator is sufficient. Theorem 8.1.4. Let A be the infinitesimal generator of the C0 -semigroup (T (t))t≥0 on the Hilbert space X. Then (T (t))t≥0 is exponentially stable if and only if (·I − A)−1 ∈ H∞ (L(X)). The definition and properties of the spaces H∞ (L(X)) and H2 (X) can be found in Appendix A.2.

8.2. Spectral projection and invariant subspaces

101

Proof. Necessity. By assumption, we know that the C0 -semigroup satisfies T (t)

≤ M eωt for some ω < 0. Proposition 5.2.4 implies that C+ 0 := {s ∈ C | Re(s) ≥ 0}, the closed right-half plane, is contained in the resolvent set of A and, furthermore, for s ∈ C+ 0 M M ≤ .

(sI − A)−1 ≤ Re(s) − ω −ω Combining this with Proposition 5.2.4.3, we conclude that (·I −A)−1 ∈ H∞ (L(X)). Sufficiency. Suppose that the C0 -semigroup satisfies T (t) ≤ M e(ω−ε)t for some positive constants M , ω and ε. It is easy to see that e−ω· T (·)x is an element of L2 ([0, ∞); X) for every x ∈ X. Furthermore, the Laplace transform of t → e−ωt T (t)x equals s → ((s+ω)I −A)−1 x (see Proposition 5.2.4). The Paley-Wiener Theorem A.2.9 implies ((· + ω)I − A)−1 x ∈ H2 (X). Now, by assumption, (·I − A)−1 ∈ H∞ (L(X)), and by Theorem A.2.10.2 it follows that (·I − A)−1 ((· + ω)I − A)−1 x ∈ H2 (X). Using the resolvent equation, we conclude that (·I − A)−1 x ∈ H2 (X), since (·I − A)−1 x = ((· + ω)I − A)−1 x + ω(·I − A)−1 ((· + ω)I − A)−1 x.

(8.10)

However, the Laplace transform of t → T (t)x is s → (sI − A)−1 x and thus by the Paley-Wiener Theorem A.2.9, we have that T (·)x ∈ L2 ([0, ∞); X). Finally, Lemma 8.1.2 shows that (T (t))t≥0 is exponentially stable.

8.2 Spectral projection and invariant subspaces In this section, we discuss various invariance concepts and study the relationships between them. First we define T (t)-invariance. Definition 8.2.1. Let V be a subspace of the Hilbert space X and let (T (t))t≥0 be a C0 -semigroup on X. We say that V is T (t)-invariant if for all t ≥ 0 T (t)V ⊂ V. Since x(t) = T (t)x0 is the solution of (8.2) we see that T (t)-invariance is equivalent to the fact that the solution stays of (8.2) in V when the initial condition lies in V . Definition 8.2.2. Let V be a subspace of the Hilbert space X and let A be an infinitesimal generator of a C0 -semigroup on X. We say that V is A-invariant if A(V ∩ D(A)) ⊂ V.

102


For finite-dimensional systems, it is well known that a (closed) subspace is T (t)-invariant if and only if it is A-invariant, where A is the infinitesimal generator of T (t) = eAt , t ≥ 0. This result generalizes to bounded generators A, see Exercise 8.4. However, this results does in general not hold for infinite-dimensional systems, see [63]. However, we do have the following partial result. Lemma 8.2.3. Let V be a closed subspace of X and let A be the infinitesimal generator of the C0 -semigroup (T (t))t≥0 . If V is T (t)-invariant, then V is Ainvariant. Proof. Let v be an arbitrary element in V ∩ D(A). Then Definition 5.2.1 implies 1 lim (T (t) − I)v = Av. t↓0 t By assumption, 1t (T (t) − I)v is an element of V for every t > 0. Thus, since V is closed, the limit is also an element of V and therefore Av ∈ V . Whether or not a C0 -semigroup has a nontrivial T (t)-invariant subspace is a fundamental question. If the spectrum of A consists of two or more regions, then this question can be answered positively, as shown in the following theorem.

Γ

σ−

σ+

Figure 8.1: Spectral decomposition Theorem 8.2.4. Let A be a closed densely defined operator on X. Assume that the spectrum of A is the union of two parts, σ + and σ − , such that a rectifiable, closed, simple curve Γ can be drawn so as to enclose an open set containing σ + in its interior and σ − in its exterior. The operator, P Γ , defined by 1 P Γx = (λI − A)−1 xdλ, (8.11) 2πi Γ


103

where Γ is traversed once in the positive direction (counterclockwise), is a projection. We call this projection the spectral projection on σ + . This projection induces a decomposition of the state space X = X + ⊕ X − , where X + = P Γ X, and X − = (I − P Γ )X.

(8.12)

Moreover, the following properties hold: 1. For x ∈ D(A) we have P Γ Ax = AP Γ x. Furthermore, for all s ∈ ρ(A) we have (sI − A)−1 P Γ = P Γ (sI − A)−1 . 2. The spaces X + and X − are A-invariant, and for all s ∈ ρ(A) there holds that (sI − A)−1 X + ⊂ X + and (sI − A)−1 X − ⊂ X − ; 3. P Γ X ⊂ D(A), and the restriction of A to X + , A+ , is a bounded operator on X +; 4. σ(A+ ) = σ + and σ(A− ) = σ − , where A− is the restriction of A to X − . Furthermore, for λ ∈ ρ(A) we have that (λI − A+ )−1 = (λI − A)−1 |X + and (λI − A− )−1 = (λI − A)−1 |X − ; 5. If σ + consists of only finitely many eigenvalues with finite order, then P Γ projects onto the space of generalized eigenvectors of the enclosed eigenvalues. Thus we have that ker(λn I − A)ν(n) = ker(λn I − A+ )ν(n) , ran P Γ = λn ∈σ+

λn ∈σ+

where ν(n) is the order of λn ; 6. If σ + = {λn } with λn an eigenvalue of multiplicity 1, then P Γ z = z, ψn φn , where φn is the eigenvector of A corresponding to λn and ψn is an eigenvector of A∗ corresponding to λn with φn , ψn = 1. The order of an isolated eigenvalue λ0 is defined as follows. We say that λ0 has order ν0 if for every x ∈ X limλ→λ0 (λ−λ0 )ν0 (λI −A)−1 x exists, but there exists an x0 such that the limit limλ→λ0 (λ− λ0 )ν0 −1 (λI − A)−1 x0 does not exist. If for every ν ∈ N there exists an xν ∈ X such that the limit limλ→λ0 (λ − λ0 )ν (λI − A)−1 xν does not exist, then the order of λ0 is infinity. An eigenvalue λ has multiplicity 1, if the order is 1, and if dim ker(λI −A) = 1. Proof. Since the mapping λ → (λI − A)−1 is uniformly bounded on Γ, P Γ is a bounded linear operator on X. For s ∈ ρ(A) we have that 1 (sI − A)−1 P Γ x = (sI − A)−1 (λI − A)−1 xdλ = P Γ (sI − A)−1 x. (8.13) 2πi Γ

104


Using the resolvent identity for the middle term we find, for s ∈ ρ(A) and outside Γ, 1 −(sI − A)−1 x (λI − A)−1 x + dλ (sI − A)−1 P Γ x = 2πi Γ s−λ s−λ 1 (λI − A)−1 x = dλ, (8.14) 2πi Γ s−λ by Cauchy’s residue theorem. Using this relation we show that P Γ is a projection. Let Γ be another rectifiable, simple, closed curve enclosing σ + counterclockwise that encircles Γ as well. Then by standard complex analysis and the fact that (sI − A)−1 is holomorphic between the two curves we have that P Γ is also given by 1 P Γx = (λI − A)−1 xdλ. 2πi Γ Hence, with (8.14) we obtain 1 (sI − A)−1 PΓ xds P ΓP Γx = 2πi Γ 1 1 (λI − A)−1 x = dλds 2πi Γ 2πi Γ s−λ 1 1 1 = ds(λI − A)−1 xdλ 2πi Γ 2πi Γ s − λ 1 = (λI − A)−1 xdλ = P Γ x, 2πi Γ

by Fubini’s theorem

where we used Cauchy’s theorem. Thus P Γ is a projection. This immediately implies that X + := P Γ X and X − := (I − P Γ )X are closed linear subspaces and X = X + ⊕ X − . Now we prove part 1 to 6. Parts 1 and 2. By (8.13) it follows that P Γ commutes with the resolvent operator, and hence also AP Γ = P Γ A on the domain of A. So X + = ran P Γ and X − = ran(I − P Γ ) are A- and (sI − A)−1 -invariant, i.e., (sI − A)−1 X + ⊂ X + , (sI − A)−1 X − ⊂ X − . Part 3. We show that P Γ X ⊂ D(A). For λ, s ∈ ρ(A) it holds that (sI − A)(λI − A)−1 = (s − λ)(λI − A)−1 + I. Therefore, for x ∈ X we obtain 1 (sI − A)−1 (s − λ)(λI − A)−1 xdλ 2πi Γ 1 1 = (sI − A)−1 (sI − A)(λI − A)−1 xdλ − xdλ 2πi Γ 2πi Γ 1 = (sI − A)−1 (sI − A)(λI − A)−1 xdλ = P Γ x. (8.15) 2πi Γ


105

(8.15) holds for any x ∈ X, and so X + = P Γ X ⊂ D(A). Since X + is A-invariant, A|X + is well-defined. A|X + is closed, since A is closed. Now A+ is defined on the whole space X + , and thus by the closed graph theorem A+ is bounded on X + . Part 4. Let s be an element of C that does not lie on Γ. We define 1 (λI − A)−1 x Qs x := dλ. 2πi Γ s−λ It is easy to see that this operator commutes with the resolvent, and similarly as in part 1 and 2 we find that Qs P Γ = P Γ Qs

and

AQs = Qs A on D(A).

(8.16)

From the first relation, we conclude that X + and X − are Qs -invariant. For s = λ we have that 1 1 (sI − A)(λI − A)−1 = (λI − A)−1 + I. s−λ s−λ Thus for x ∈ D(A) and s ∈ /Γ 1 1 (sI − A)(λI − A)−1 x 1 (sI −A)Qs x = dλ = P Γ x+ xdλ. (8.17) 2πi Γ s−λ 2πi Γ s − λ For s outside Γ and x ∈ X + , we get (sI − A+ )Qs x = (sI − A)Qs x = P Γ x = x, where in the first equality we have used that Qs maps X + into X + . Since Qs commutes with A, we find (sI − A+ )Qs = Qs (sI − A+ ) = IX + .

(8.18)

Similar, we find, for s inside Γ, (sI−A− )Qs = IX − and Qs (sI−A− )x = x

for x ∈ D(A− ) = D(A)∩X − . (8.19)

Thus any s outside Γ lies in the resolvent set of A+ , and any s inside Γ lies in the resolvent set of A− . On the other hand, the A- and (sI − A)−1 -invariance of X + and X − , see part 1 and 2, gives that ρ(A) = ρ(A+ ) ∩ ρ(A− ), and (equivalently) σ(A) = σ(A+ ) ∪ σ(A− ).

(8.20)

Indeed if λ ∈ ρ(A) then by part 2 (λI − A)−1 maps X + into X + . Thus (λIX + − A+ )(λI − A)−1 |X + is well defined and (λIX + − A+ )(λI − A)−1 |X + = (λI − A)(λI − A)−1 |X + = IX + .

106


On the other hand, on X + we have that (λI − A)−1 |X + (λIX + − A+ ) = (λI − A)−1 |X + (λI − A)|X + = (λI − A)−1 (λI − A)|X + = IX + .

(by part 2)

So ρ(A) ⊂ ρ(A+ ) and (λI − A)−1 |X + = (λIX + − A+ )−1 . Similarly, we can show that ρ(A) ⊂ ρ(A− ) and (λI − A)−1 |X − = (λIX − − A− )−1 . Thus we deduce that ρ(A) ⊂ ρ(A+ ) ∩ ρ(A− ). If s ∈ ρ(A+ )∩ρ(A− ), then by using the following decomposition for x ∈ D(A), (sI − A)x = (sI − A+ )P Γ x + (sI − A− )(I − P Γ )x,

(8.21)

it is easy to see that (sI − A+ )−1 P Γ + (sI − A− )−1 (I − P Γ ) is the inverse of (sI − A). Thus (8.20) is proved. By definition, we have that σ(A) = σ + ∪ σ − . Furthermore, we showed that + σ ⊂ ρ(A− ) and σ − ⊂ ρ(A+ ). Combining this with (8.20) we find σ(A+ ) = σ + and σ(A− ) = σ − . Part 5. We may write σ + = {λ1 , λ2 , . . . , λN }. By standard complex integration theory we have that 1 P Γx = (λI − A)−1 xdλ 2πi Γ N N 1 = (λI − A)−1 xdλ = P Γn x, 2πi Γn n=1 n=1 where Γn is a rectifiable, closed, simple curve enclosing only λn . Thus it suffices to prove the assertion for the case that σ + = {λn }, where λn is an eigenvalue with finite order ν(n). We do this for the generic case ν(n) = 1; the general case can be found in [10]. Let Γ denote the rectifiable, simple, closed curve that encloses counterclockwise only one point λn in the spectrum of A. First we prove that ran P Γ ⊂ ker(λn I − A). For x ∈ X and s ∈ ρ(A) we have 1 (λn − s)P Γ x = (λn − s)(λI − A)−1 xdλ. (8.22) 2πi Γ Multiplying (8.15) from the left by (sI − A) and adding (8.22) yields 1 (λn I − A)P Γ x = (λn − λ)(λI − A)−1 xdλ. 2πi Γ This last expression is zero, since (λn − ·)(·I − A)−1 x is holomorphic inside Γ. This proves ran P Γ ⊂ ker(λn I −A); to prove the other inclusion note that (λn I −A)x0 = 1 x0 . This implies 0 implies that (λI − A)−1 x0 = λ−λ n 1 1 1 P Γ x0 = (λI − A)−1 x0 dλ = x0 dλ = x0 . 2πi Γ 2πi Γ λ − λn


107

Part 6. In part 5 we showed that P Γ maps onto the span of φn . Hence P Γ z = h(z)φn , where h is a function from X to C. Since P Γ is a bounded linear operator, it follows that h is an element of L(X, C). From the Riesz Representation Theorem it follows that h(z) = z, ψn for some ψn ∈ X. Consider now for x ∈ D(A), x, A∗ ψn φn = Ax, ψn φn = P Γ Ax = AP Γ x = Ax, ψn φn = x, λn ψn φn . Since this holds for every x ∈ D(A), we conclude that A∗ ψn = λn ψn . Furthermore, using the fact that P Γ φn = φn it follows easily that φn , ψn = 1. Theorem 8.2.5. Let A be the infinitesimal generator of the C0 -semigroup (T (t))t≥0 on X. Assume that the spectrum of A is the union of two parts, σ + and σ − as in Theorem 8.2.4. Then X + and X − are T (t)-invariant and (T + (t))t≥0 , (T − (t))t≥0 with T + (t) = T (t)|X + , T − (t) = T (t)|X − define C0 -semigroups on X + and X − , respectively. The infinitesimal generator of (T + (t))t≥0 is A+ , whereas A− is the infinitesimal generator of (T − (t))t≥0 . Proof. Since (λI − A)−1 T (t) = T (t)(λI − A)−1 for all λ ∈ ρ(A) and t ≥ 0, it follows that P Γ commutes with T (t). Then it is easily proved that X + and X − are T (t)-invariant. Consequently, (T + (t))t≥0 and (T − (t))t≥0 with T + (t) = T (t)|X + , T − (t) = T (t)|X − are C0 -semigroups on X + and X − , respectively. We shall only prove that A− is the infinitesimal generator of (T − (t))t≥0 , as − exists for the proof for (T + (t))t≥0 is very similar. Suppose that lim T (t)x−x t t→0+

x ∈ X − . Since T − (t)x = T (t)x we conclude that x ∈ D(A) and hence x is an element of D(A− ). By definition, the limit equals Ax = A− x. On the other hand, if x ∈ D(A− ), then x ∈ D(A) and so the limit exists and equals A− x. Combining these results the infinitesimal generator of (T − (t))t≥0 is A− . We combine the previous theorem with Theorem 8.1.4 to find a sufficient condition guaranteeing that (T − (t))t≥0 is exponentially stable. Lemma 8.2.6. Let A be the infinitesimal generator of the C0 -semigroup (T (t))t≥0 on X. Assume that the spectrum of A is the union of two parts, σ + and σ − , such − that σ + ⊂ C+ ⊂ C− 0 = {s ∈ C | Re(s) ≥ 0}, σ 0 = {s ∈ C | Re(s) < 0}, and that a rectifiable, closed, simple curve Γ can be drawn so as to enclose an open set containing σ + in its interior and σ − in its exterior. Assume further that

(sI − A)−1 < ∞.

sup s∈C+ 0 ,s

outside

(8.23)

Γ

Then the semigroup (T − (t))t≥0 := (T (t)|X − )t≥0 is exponentially stable, where X − is the subspace of X associated to σ − , see (8.12).

108


Proof. By Theorem 8.2.5 the infinitesimal generator of (T (t)|X − )t≥0 equals A− . The spectrum of A− lies outside Γ, see Theorem 8.2.4.4. Since the interior of Γ is a bounded set, and since the resolvent is holomorphic on the resolvent set, we find that (8.24) sup (sI − A− )−1 < ∞. s −

Furthermore, for x ∈ X

−

inside

Γ

and s ∈ C+ 0 outside Γ, we have that

(sI − A− )−1 x− = (sI − A)−1 x− . Combining this with (8.23) and (8.24) gives that the resolvent operator of A− is − uniformly bounded on C+ 0 , and by Theorem 8.1.4 we conclude that (T (t))t≥0 is exponentially stable.

8.3 Exercises 8.1. In this exercise we introduce a different concept of stability, and show that it is weaker than exponential stability. Definition 8.3.1. The C0 -semigroup (T (t))t≥0 on the Hilbert space X is strongly stable if for every x ∈ X, T (t)x converges to zero as t tends to ∞. Let X be the Hilbert space L2 (0, ∞) and let the operators T (t) : X → X, t ≥ 0, be defined by (T (t)f )(ζ) := f (t + ζ). a) Show that (T (t))t≥0 is a C0 -semigroup on X. b) Prove that (T (t))t≥0 is strongly stable, but not exponentially stable. 8.2. A natural question is whether the following condition is sufficient for strong or exponential stability, Ax, x + x, Ax < 0,

for all x ∈ D(A) \ {0}.

(8.25)

In this exercise we show that this does not hold in general. Let X be the Hilbert space L2 (0, ∞) equipped with the inner product ∞ f, g := f (ζ)g(ζ)(e−ζ + 1)dζ, 0

and let the operators T (t) : X → X, t ≥ 0, be defined by " f (ζ − t), ζ > t, (T (t)f )(ζ) := 0, 0 ≤ ζ < t.


109

a) Show that (T (t))t≥0 is a C0 -semigroup on X. b) Prove that (T (t))t≥0 is not strongly stable. c) Prove T (t2 )x < T (t1 )x for all x ∈ X \ {0} and t2 > t1 ≥ 0. d) Show that the infinitesimal generator of (T (t))t≥0 is given by Af = −

df dζ

with domain D(A) = {f ∈ X | f is absolutely continuous,

df ∈ X, and f (0) = 0}. dζ

Hint: See Exercise 6.3.b. e) Use the previous item to prove that (8.25) holds. 8.3. Let Q be a bounded, self-adjoint operator on the Hilbert space X which satisfies Q ≥ mI, for some m > 0. Let A be the infinitesimal generator of a strongly continuous semigroup on X which satisfies the Lyapunov inequality Ax, Qx + Qx, Ax ≤ 0,

x ∈ D(A).

Show that there exists an equivalent norm on X such that A generates a contraction semigroup with respect to this new norm. 8.4. Let A be a bounded operator on the Hilbert space X, and let V be a closed linear subspace of X. Show that V is T (t)-invariant if and only if V is Ainvariant. 8.5. Let A be the infinitesimal generator of the C0 -semigroup (T (t))t≥0 and let V be a one-dimensional linear subspace of X. Show that V is T (t)-invariant if and only if V = span{φ} with φ an eigenvector of A. Hint: You may use the following fact: If a continuous scalar function f satisfies f (t + s) = f (t)f (s) for t, s ∈ (0, ∞), then f (t) = eλt for some λ.

8.4 Notes and references Exponential stability is one of the most studied properties of semigroups, and therefore the results in the first section can be found in many books, we refer to chapter 5 of Curtain and Zwart [10] or chapter V of Engel and Nagel [15]. The characterization as presented in Theorem 8.1.4 originates from Huang [26]. The relation between T (t)- and A-invariance is studied in [45], [63], see also [10]. Spectral projections play an important role in many subfields of functional analysis, and thus the results as formulated in Theorem 8.2.4 can be found at many places, see [10, Lemma 2.5.7] or [18, Theorem XV.2.1].

Chapter 9

Stability of Port-Hamiltonian Systems In this chapter we return to the class of port-Hamiltonian partial differential equations which we introduced in Chapter 7. If a port-Hamiltonian system possesses n (linearly independent) boundary conditions and if the energy is non-increasing, then the associated abstract differential operator generates a contraction semigroup on the energy space. This chapter is devoted to exponential stability of port-Hamiltonian systems. Exercise 8.2 implies that the condition Ax, x + x, Ax < 0,

x ∈ D(A), x = 0,

(9.1)

is in general not sufficient for strong stability of the semigroup generated by A. However, in this chapter we show that if a weaker but more structured condition than (9.1) holds with respect to the energy inner product for a port-Hamiltonian system, then the port-Hamiltonian system is even exponentially stable, see Theorem 9.1.3.

9.1 Exponential stability of port-Hamiltonian systems We start with a repetition of the definition of homogeneous port-Hamiltonian systems as introduced in Section 7.1. Moreover, we equip the port-Hamiltonian system with boundary conditions, that is we study differential equations of the form ∂ ∂x (ζ, t) = P1 (H(ζ)x(ζ, t)) + P0 (H(ζ)x(ζ, t)) (9.2) ∂t ∂ζ with the boundary condition WB

f∂ (t) e∂ (t)

= 0,

B. Jacob and H.J. Zwart, Linear Port-Hamiltonian Systems on Infinite-dimensional Spaces, Operator Theory: Advances and Applications 223, DOI 10.1007/978-3-0348-0399-1_9, © Springer Basel 2012

(9.3)

111

112

Chapter 9. Stability of Port-Hamiltonian Systems

where

f∂ (t) e∂ (t)

1 P1 = √ 2 I

−P1 I

H(b)x(b, t) H(a)x(a, t)

.

(9.4)

We assume that the following conditions hold, see Definition 7.1.2 and Theorem 7.2.4. Assumption 9.1.1. • P1 ∈ Kn×n is invertible and self-adjoint; • P0 ∈ Kn×n is skew-adjoint; • H ∈ C 1 ([a, b]; Kn×n ), H(ζ) is self-adjoint for all ζ ∈ [a, b] and mI ≤ H(ζ) ≤ M I for all ζ ∈ [a, b] and some M, m > 0 independent of ζ; • WB ∈ Kn×2n has full rank; • WB ΣWB∗ ≥ 0, where Σ = [ I0 I0 ]. We note that Assumption 9.1.1 is slightly stronger than the assumptions used in Chapter 7 as we did not assume that H is continuous differentiable in Chapter 7. However, we would like to remark that our main theorem (Theorem 9.1.3) also holds if P0 satisfies P0 + P0∗ ≤ 0. Throughout this section we assume that Assumption 9.1.1 is satisfied. Theorem 7.2.4 implies that the operator A given by d (9.5) Ax := P1 (Hx) + P0 (Hx) dζ with domain D(A) = {x ∈ L2 ([a, b]; Kn ) | Hx ∈ H 1 ([a, b]; Kn ), WB

f

∂

e∂

= 0}

(9.6)

generates a contraction semigroup on the state space X = L2 ([a, b]; Kn )

(9.7)

equipped with the inner product f, g X

1 = 2

b

g(ζ)∗ H(ζ)f (ζ)dζ.

(9.8)

a

The associated norm is denoted by · X . In the following lemma, we show that the norm/energy of a state trajectory can be bounded by the energy at one of the boundaries. This result is essential for the proof of exponential stability of port-Hamiltonian systems. The proof is based on an idea of Cox and Zuazua in [8].

9.1. Exponential stability of port-Hamiltonian systems

113

Lemma 9.1.2. Let A be defined by (9.5) and (9.6), and let (T (t))t≥0 be the contraction semigroup generated by A. Then there exists constants τ > 0 and c > 0 such that for every x0 ∈ D(A) the state trajectory x(t) := T (t)x0 satisfies τ

x(τ ) 2X ≤ c

H(b)x(b, t) 2 dt and (9.9) 0 τ

x(τ ) 2X ≤ c

H(a)x(a, t) 2 dt. (9.10) 0

Proof. Let x0 ∈ D(A) be arbitrary and let x(t) := T (t)x0 , t ≥ 0. x0 ∈ D(A) implies that x(t) ∈ D(A) for all t. Furthermore, x is the classical solution of (9.2) with x(0) = x0 . In the following we write x(ζ, t) instead of (x(t))(ζ). For this trajectory, we define the function F : [a, b] → R by

τ −γ(b−ζ)

F (ζ) =

x∗ (ζ, t)H(ζ)x(ζ, t) dt,

ζ ∈ [a, b],

(9.11)

γ(b−ζ)

where we assume that γ > 0 and τ > 2γ(b − a). The second condition implies in particular τ − γ(b − ζ) > γ(b − ζ), that is, we are not integrating over a negative time interval. Differentiating the function F gives dF (ζ) = dζ

τ −γ(b−ζ)

∂ (H(ζ)x(ζ, t)) dt ∂ζ γ(b−ζ) ∗ τ −γ(b−ζ) ∂ + x(ζ, t) H(ζ)x(ζ, t) dt ∂ζ γ(b−ζ) x∗ (ζ, t)

+ γx∗ (ζ, γ(b − ζ))H(ζ)x(ζ, γ(b − ζ)) + γx∗ (ζ, τ − γ(b − ζ))H(ζ)x(ζ, τ − γ(b − ζ)). Since P1 is invertible and since x satisfies (9.2), we obtain (for simplicity we omit the dependence on ζ and t) dF (ζ) = dζ

τ −γ(b−ζ)

∗

x

P1−1

γ(b−ζ)

τ −γ(b−ζ)

+ γ(b−ζ)

P1−1

∂x − P0 Hx dt ∂t

∗ ∂x dH − x − P1−1 P0 Hx x dt ∂t dζ

+ γx∗ (ζ, τ − γ(b − ζ))H(ζ)x(ζ, τ − γ(b − ζ)) + γx∗ (ζ, γ(b − ζ))H(ζ)x(ζ, γ(b − ζ)) τ −γ(b−ζ) τ −γ(b−ζ) ∂x ∗ −1 dH ∗ −1 ∂x + x dt = x P1 P1 x dt − x∗ ∂t ∂t dζ γ(b−ζ) γ(b−ζ) τ −γ(b−ζ) − x∗ HP0∗ P1−1 + P1−1 P0 H x dt γ(b−ζ)

114

Chapter 9. Stability of Port-Hamiltonian Systems + γx∗ (ζ, τ − γ(b − ζ))H(ζ)x(ζ, τ − γ(b − ζ)) + γx∗ (ζ, γ(b − ζ))H(ζ)x(ζ, γ(b − ζ))

where we have used that P1∗ = P1 , H∗ = H. The first integral can be calculated, and so we find τ −γ(b−ζ) t=τ −γ(b−ζ) dF dH −1 ∗ (ζ) = x (ζ, t)P1 x(ζ, t) t=γ(b−ζ) − x∗ x dt dζ dζ γ(b−ζ) τ −γ(b−ζ) − x∗ HP0∗ P1−1 + P1−1 P0 H x dt γ(b−ζ)

+ γx∗ (ζ, τ − γ(b − ζ))H(ζ)x(ζ, τ − γ(b − ζ)) + γx∗ (ζ, γ(b − ζ))H(ζ)x(ζ, γ(b − ζ)) τ −γ(b−ζ) dH −1 ∗ ∗ −1 =− x HP0 P1 + P1 P0 H + x dt dζ γ(b−ζ) + x∗ (ζ, τ − γ(b − ζ)) P1−1 + γH(ζ) x(ζ, τ − γ(b − ζ)) + x∗ (ζ, γ(b − ζ)) −P1−1 + γH(ζ) x(ζ, γ(b − ζ)). We now choose γ large enough, such that P1−1 + γH and −P1−1 + γH are coercive (positive definite). Note that τ > 2γ(b − a), and thus a large constant γ implies that τ must be large as well. Using the coercivity of P1−1 + γH and −P1−1 + γH, we find that τ −γ(b−ζ) dF dH −1 ∗ ∗ −1 (ζ) ≥ − x HP0 P1 + P1 P0 H + x dt. dζ dζ γ(b−ζ) Since P1 and P0 are constant matrices and, by assumption, dH dζ (ζ) is bounded, there exists a constant κ > 0 such that for all ζ ∈ [a, b] we have H(ζ)P0∗ P1−1 + P1−1 P0 H(ζ) +

dH ≤ κH(ζ). dζ

Thus, for all ζ ∈ [a, b] we obtain τ −γ(b−ζ) dF (ζ) ≥ −κ x∗ (ζ, t)H(ζ)x(ζ, t) dt = −κ F (ζ), dζ γ(b−ζ)

(9.12)

where we used (9.11). For simplicity, we denote the derivative of F by F . Inequality (9.12) implies, for all ζ1 ∈ [a, b], b b F (ζ) dζ ≥ −κ 1 dζ, (9.13) ζ1 F (ζ) ζ1 or equivalently

ln F (b) − ln F (ζ1 ) ≥ −κ (b − ζ1 ).

(9.14)


115

Thus we obtain F (b) ≥ F (ζ1 ) e−κ (b−ζ1 ) ≥ F (ζ1 ) e−κ (b−a) for ζ1 ∈ [a, b].

(9.15)

On the other hand, since x(t2 ) X ≤ x(t1 ) X for any t2 ≥ t1 (by the contraction property of the semigroup), we deduce that τ −γ(b−a) τ −γ(b−a)

x(t) 2X dt ≥ x(τ − γ(b − a)) 2X 1 dt γ(b−a)

γ(b−a)

= (τ − 2γ(b − a)) x(τ − γ(b − a)) 2X . Using the definition of F and x(t) 2X , see (9.11) and (9.8), together with the equation above, the estimate (9.15), and the coercivity of H we obtain 2(τ − 2γ(b − a)) x(τ ) 2X ≤ 2(τ − 2γ(b − a)) x(τ − γ(b − a)) 2X τ −γ(b−a) ≤2

x(t) 2X dt γ(b−a)

τ −γ(b−a) b

=

γ(b−a) b

τ −γ(b−a)

= a

x∗ (ζ, t)H(ζ)x(ζ, t) dζ dt

a

x∗ (ζ, t)H(ζ)x(ζ, t) dt dζ

γ(b−a)

(by Fubini’s theorem) b τ −γ(b−ζ) ≤ x∗ (ζ, t)H(ζ)x(ζ, t) dt dζ

a

γ(b−ζ)

(since the integration interval increases) b

F (ζ) dζ ≤ (b − a)F (b) eκ (b−a) τ κ (b−a) = (b − a) e x∗ (b, t)H(b)x(b, t) dt 0 τ −1 κ (b−a) ≤ m (b − a) e

H(b)x(b, t) 2 dt. =

a

0

Hence for our choice of τ we have that τ 2

x(τ ) X ≤ c

H(b)x(b, t) 2 dt,

(9.16)

0 κ (b−a)

(b−a) e where c = 2(τ . This proves estimate (9.9). −2γ(b−a))m The second estimate follows in a similar manner by replacing the function F in the calculations above by τ −γ(ζ−a) F˜ (ζ) = x∗ (ζ, t)H(ζ)x(ζ, t) dt. γ(ζ−a)

116

Chapter 9. Stability of Port-Hamiltonian Systems Using this technical lemma the proof of exponential stability is easy.

Theorem 9.1.3. Let A be defined by (9.5) and (9.6). If for some positive constant k one of the following conditions is satisfied for all x0 ∈ D(A), Ax0 , x0 X + x0 , Ax0 X ≤ −k H(b)x0 (b) 2

(9.17)

Ax0 , x0 X + x0 , Ax0 X ≤ −k H(a)x0 (a) ,

(9.18)

2

then A generates an exponentially stable C0 -semigroup. Proof. Without loss of generality we assume that the first inequality (9.17) holds. By Lemma 9.1.2 there exist positive constants τ and c such that (9.9) holds. Let x0 ∈ D(A) and by (T (t))t≥0 we denote the C0 -semigroup generated by A. We define x(t) := T (t)x0 , t ≥ 0. x0 ∈ D(A) implies x(t) ∈ D(A) and x(t) ˙ = Ax(t), t ≥ 0. Using this differential equation, we obtain dx(t), x(t) X d x(t) 2X = = Ax(t), x(t) X + x(t)Ax(t) X . dt dt

(9.19)

Equations (9.17) and (9.19) now imply that τ d x(t) 2X

x(τ ) 2X − x(0) 2X = dt dt 0 τ ≤ −k

H(b)x(b, t) 2 dt. 0

Combining this with (9.9), we find that

x(τ ) 2X − x(0) 2X ≤

−k

x(τ ) 2X . c

c Thus x(τ ) 2X ≤ c+k

x(0) 2X , which implies that the semigroup (T (t))t≥0 generated by A satisfies T (τ ) < 1, and therefore (T (t))t≥0 is exponentially stable by Theorem 5.1.5.

Estimate (9.17) provides a simple test for exponential stability of port-Hamiltonian systems. We note that Lemma 7.2.1 and (7.23) imply 1 ((H(b)x(b))∗ P1 H(b)x(b) − (H(a)x(a))∗ P1 H(a)x(a)) 2 1 = (f∂∗ e∂ + e∗∂ f∂ ) . (9.20) 2

Ax, x X + x, Ax X =

This equality can be used to establish (9.17) or (9.18). An even simpler sufficient condition is provided by the following lemma. Lemma 9.1.4. Consider the port-Hamiltonian system (9.2)–(9.3). If WB ΣWB∗ > 0, then the port-Hamiltonian system is exponentially stable.


117

Proof. Lemma 7.3.1 implies that WB can be written as WB = S I + V I − V , where S, V ∈ Kn×n , S is invertible and V V ∗ < I. We define WC = I + V ∗ −I + V ∗ . It is easy to see that with this choice, I+V I−V

WC is a n×2n matrix with rank n. Next we prove that the matrix I+V is ∗ ∗ −I+V I+V I−V

x y = invertible. Assuming I+V ∗ −I+V ∗ is not invertible, there exists a vector

I+V I−V

∗ 0 such that x y I+V = 0, or equivalently (I + V )x + (I + V )y = 0 ∗ ∗ −I+V ∗ and (I − V )x + (−I + V ∗ )y = 0. These equations together with V V < I imply I+V I−V

is invertible. This implies the x = y = 0, and thus the matrix I+V ∗ −I+V ∗ WB

invertibility of the matrix WC . f Let x ∈ D(A) be arbitrary. The definition of the domain of A implies that ∂ e∂ ∈ ker WB . Lemma 7.3.2 implies that f∂ I −V = (9.21) e∂ −I − V for some ∈ Kn . Thus with (9.20) we find Ax, x X + x, Ax X = ∗ (−I + V ∗ V ) . Furthermore, we have the equality

f∂ I −V ∗ ∗ = I + V , −I + V = 2(I − V ∗ V ) . y := WC −I − V e∂

(9.22)

(9.23)

Combining equations (9.22) and (9.23) we obtain Ax, x X + x, Ax X =

1 ∗ y [−I + V ∗ V ]−1 y ≤ −m1 y 2 4

(9.24)

for some m1 > 0. Here we have used that V V ∗ < I, or equivalently V ∗ V − I < 0. Using (9.4), the boundary condition, and the definition of y the relation between y and x is given by 1 0 WB P1 −P1 H(b)x(b) H(b)x(b) =√ =: W . y I I H(a)x(a) H(a)x(a) 2 WC B

Since P1 and W WC are invertible, it follows that the matrix W is invertible and, in particular, W w 2 ≥ m2 w 2 for every w ∈ Kn and some m2 > 0. Taking norms on both sides yields 2 2 H(b)x(b) ≥ m2 H(b)x(b) ≥ m2 H(b)x(b) 2 .

y 2 = W (9.25) H(a)x(a) H(a)x(a) Combining the estimates (9.24) and (9.25) we obtain Ax, x X + x, Ax X ≤ −m1 y 2 ≤ −m1 m2 (Hx)(b) 2 . Thus (9.17) holds, and therefore Theorem 9.1.3 implies the exponential stability of the port-Hamiltonian system.

118


Unfortunately, the sufficient condition of the previous lemma often cannot be applied, as the condition implies that the system possesses as many dampers as boundary controls. In practice fewer dampers are necessary as it is shown in the example of the following section.

9.2 An example In this section we show how to apply the results of the previous section to the vibrating string. Example 9.2.1. Consider the vibrating string on the spatial interval [a, b]. In Example 7.1.1 we saw that the model is given by ∂2w 1 ∂ ∂w (ζ, t) = T (ζ) (ζ, t) , (9.26) ∂t2 ρ(ζ) ∂ζ ∂ζ where ζ ∈ [a, b] is the spatial variable, w(ζ, t) is the vertical position of the string at position ζ and time t, T is the Young’s modulus of the string, and ρ is the mass density. This system has the energy/Hamiltonian 2 2 1 b ∂w ∂w E(t) = ρ(ζ) (ζ, t) + T (ζ) (ζ, t) dζ. (9.27) 2 a ∂t ∂ζ

11 00 00 11 00 11 00 11 00 11

As depicted in Figure 9.1, the string is fixed at the left-hand side and at

Figure 9.1: The vibrating string with a damper the right-hand side we attach a damper, that is, the force at this tip equals a (negative) constant times the velocity of the tip, i.e, T (b)

∂w ∂w (b, t) = −k (b, t), ∂ζ ∂t

k ≥ 0.

(9.28)

For this example the boundary effort and boundary flow are given by, see Example 7.2.5, ∂w ∂w ∂w 1 T (b) ∂w 1 ∂ζ (b) − T (a) ∂ζ (a) ∂t (b) + ∂t (a) √ f∂ = √ , e = . ∂ ∂w ∂w ∂w ∂w 2 2 T (b) ∂ζ (b) + T (a) ∂ζ (a) ∂t (b) − ∂t (a) (9.29)

9.2. An example

119

The boundary conditions

with WB =

√1 2

0 0

∂w ∂t (a)

= 0 and (9.28) can be equivalently written as

(b) + k ∂w (b) T (b) ∂w ∂ζ ∂t ∂w ∂t (a) 1 1 k k 1 f∂ f∂ = √ = WB , e∂ e∂ 2 0 −1 1 0 =

1

k k 1 0 −1 1 0

(9.30)

. WB is a 2 × 4-matrix with rank 2, and WB ΣWB∗

=

2k 0

0 0

.

Since the matrix WB ΣWB∗ is non-negative, the infinitesimal generator associated to this p.d.e., generates a contractions semigroup. However, WB ΣWB∗ is not positive definite, and so we cannot used Lemma 9.1.4 to conclude that the corresponding semigroup is exponentially stable. We will show that (9.17) holds. ∂w ρ ∂t As state variables we have chosen x = ∂w and H is given by H(ζ) = ∂ζ 1 0 ρ(ζ) . By equation (9.20) we have that 0 T (ζ) ∂w ∂w ∂w ∂w (b)T (b) (b) − (a)T (a) (a) ∂t ∂ζ ∂t ∂ζ 2 ∂w = −k (b) , ∂t

Ax, x X + x, Ax X =

(9.31)

where we have used the boundary conditions. Moreover, we calculate

H(b)x(b) 2 =

2 2 2 ∂w ∂w ∂w (b) + T (b) (b) = (k 2 + 1) (b) . ∂t ∂ζ ∂t

(9.32)

Combining the two previous equations, we find that Ax, x X + x, Ax X ≤ −

k

H(b)x(b) 2 . 1 + k2

(9.33)

Hence using Theorem 9.1.3 we may conclude that attaching a damper at one end of the transmission line stabilizes the system exponentially. We remark that if the damper is not connected, i.e., k = 0, then the corresponding semigroup is a unitary group, and the solution cannot be exponentially stable.

120


9.3 Exercises 9.1. Consider the transmission line on the spatial interval [a, b] as discussed in Exercise 7.1 ∂Q ∂ (ζ, t) = − ∂t ∂ζ ∂φ ∂ (ζ, t) = − ∂t ∂ζ

φ(ζ, t) , L(ζ) Q(ζ, t) . C(ζ)

(9.34)

Here Q(ζ, t) is the charge at position ζ ∈ [a, b] and time t > 0, and φ(ζ, t) is the (magnetic) flux at position ζ and time t. C is the (distributed) capacity and L is the (distributed) inductance. V (a)

V (b)

I(a) b

a

I(b)

Figure 9.2: Transmission line The voltage and current are given by V = Q/C and I = φ/L, respectively. We set the voltage at ζ = a to zero, and put a resistor at the other end. This implies that we have the p.d.e. (9.34) with boundary conditions V (a, t) = 0,

V (b, t) = RI(b, t),

(9.35)

with R > 0. (a) Show that the differential operator associated to the p.d.e. (9.34) with boundary conditions (9.35) generates a contraction semigroup on the energy space, see (7.53). (b) Prove that the semigroup as defined in the previous item is exponentially stable. 9.2. Consider a flexible beam modeled by the Timoshenko beam equations, see Example 7.1.4. We assume that the beam is clamped at the left-hand side, i.e., at ζ = a, and at the right-hand side we apply a damping force proportional to the velocity. Thus the boundary conditions are ∂w (a, t) = 0, ∂t

∂φ (a, t) = 0, ∂t

and K(b)

EI(b)

∂φ ∂φ (b, t) = −α1 (b, t) ∂ζ ∂t

∂w ∂w (b, t) − φ(b, t) = −α2 (b, t). ∂ζ ∂t

(9.36)

(9.37)

9.3. Exercises

121

(a) Assume that α1 , α2 ∈ [0, ∞). Show that under these conditions the Timoshenko beam associated with boundary conditions (9.36) and (9.37) generates a contraction semigroup on its energy space. (b) Prove that the semigroup as defined in the previous item is exponentially stable if α1 and α2 are positive. (c) Show that the semigroup as defined in the first item is not exponentially stable if α1 = 0 and α2 = 0. 9.3. Consider coupled vibrating strings as given in the figure below. We assume

I

1 0 0 1 0 1 0 1 0 1 0 1

II

III

Figure 9.3: Coupled vibrating strings with dampers that the length of all strings are equal. The model for every vibrating string is given by (9.26) with physical parameters, ρI , TI , ρII , etc. Furthermore, we assume that the three strings are connected via a (mass-less) bar, as shown in Figure 9.3. This bar can only move in the vertical direction. This implies I that the velocity of string I, ∂w , at its right-hand side equals those of the ∂t other two strings at their left-hand side. Furthermore, the force of string I at its right-end side equals the sum of the forces of the other two at their left-hand side. We assume that for all three waves, the connecting point is at coordinate ζ = a. Thus the balance of forces in the middle is given by −TI (a)

∂wI ∂wII ∂wIII (a) = TII (a) (a) + TIII (a) (a). ∂ζ ∂ζ ∂ζ

(9.38)

As depicted, at the outer points dampers are attached. Thus Tk (b)

∂wk ∂wk (b) = −αk (b), ∂ζ ∂t

k = I, II, and III.

(9.39)

(a) Assume that the constants αI , αII and αIII are non-negative. Show that then the coupled wave equation of Figure 9.3 generates a contraction semigroup on its energy space.

122

Chapter 9. Stability of Port-Hamiltonian Systems (b) Prove that the semigroup as defined in the previous item is exponentially stable if αI , αII , and αIII are positive.

9.4 Notes and references Exponential stability of a system described by a partial differential equation is a well-studied topic. There are numerous papers on this subject, using different techniques. In the appendix of Cox and Zuazua [8], a simple proof is presented showing that the wave equation with damping at one boundary as in Example 9.2.1 is exponentially stable. It is shown in [59] that this idea can be applied to all (first order) port-Hamiltonian systems. The result as presented here follows [59].

Chapter 10

Inhomogeneous Abstract Differential Equations and Stabilization In the previous five chapters we considered the homogeneous (abstract) differential equation x(t) ˙ = Ax(t),

x(0) = x0 .

(10.1)

However, for control theoretical questions it is essential to add an input to the differential equation, see e.g. Chapters 3 and 4. Section 10.1 is devoted to infinite-dimensional inhomogeneous differential equations and in Section 10.2 we add an output equation. The obtained formulas will be very similar to those found in Chapter 2. However, C0 -semigroups are in general not differentiable on an infinite-dimensional state space and thus the proofs are more involved. As a control application we study in Section 10.4 the stabilization of an infinite-dimensional system by a finite-dimensional feedback. This result generalizes Theorem 4.3.3 to an infinite-dimensional state space. Since for stabilization of a finite-dimensional system we chose to work on Cn , we assume throughout this chapter that our state space X is a Hilbert space over the complex numbers.

10.1 The abstract inhomogeneous Cauchy problem If A is the infinitesimal generator of a C0 -semigroup (T (t))t≥0 , then the classical solution of the abstract homogeneous Cauchy initial value problem x(t) ˙ = Ax(t), t ≥ 0,

x(0) = x0 ∈ D(A)

B. Jacob and H.J. Zwart, Linear Port-Hamiltonian Systems on Infinite-dimensional Spaces, Operator Theory: Advances and Applications 223, DOI 10.1007/978-3-0348-0399-1_10, © Springer Basel 2012

123

124

Chapter 10. Inhomogeneous Abstract Differential Equations

is given by x(t) = T (t)x0 , see Lemma 5.3.2. In this section we consider the abstract inhomogeneous Cauchy problem x(t) ˙ = Ax(t) + f (t), t ≥ 0,

x(0) = x0 ,

(10.2)

where for the moment we assume that f ∈ C([0, τ ]; X). (10.2) is also called an abstract evolution equation or abstract differential equation. First we define what we mean by a solution of (10.2), and we begin with the notion of a classical solution. C 1 ([0, τ ]; X) denotes the class of continuous functions on [0, τ ] whose derivative is again continuous on [0, τ ]. Definition 10.1.1. Consider equation (10.2) on the Hilbert space X and let τ > 0. The function x : [0, τ ] → X is a classical solution of (10.2) on [0, τ ] if x ∈ C 1 ([0, τ ]; X), x(t) ∈ D(A) for all t ∈ [0, τ ] and x(t) satisfies (10.2) for all t ∈ [0, τ ]. The function x is a classical solution on [0, ∞) if x is a classical solution on [0, τ ] for every τ ≥ 0. In the following we extend the finite-dimensional results of Section 2.2 to the infinite-dimensional situation. Lemma 10.1.2. Assume that f ∈ C([0, τ ]; X) and that x : [0, τ ] → X is a classical solution of (10.2) on [0, τ ]. Then Ax(·) is an element of C([0, τ ]; X), and we have

t

T (t − s)f (s)ds,

x(t) = T (t)x0 +

t ∈ [0, τ ].

(10.3)

0

Proof. From (10.2), we may conclude Ax(t) = x(t) ˙ − f (t) and x˙ ∈ C([0, τ ]; X) implies Ax(·) ∈ C([0, τ ]; X). Next we prove (10.3). Let t be an arbitrary, but fixed, element of [0, τ ] and consider the function s → T (t−s)x(s). We show that this function is differentiable on [0, t). Let h be sufficiently small and consider T (t − s − h)x(s + h) − T (t − s)x(s) T (t − s − h)x(s + h) − T (t − s − h)x(s) = h h T (t − s − h)x(s) − T (t − s)x(s) + . h If h converges to zero, then the second term on the right hand side converges to −AT (t − s)x(s), since x(s) ∈ D(A). Thus it remains to show that the first term on the right hand side converges. We have the equality T (t − s − h)x(s + h) − T (t − s − h)x(s) − T (t − s)x(s) ˙ h x(s + h) − x(s) = T (t − s − h) − x(s) ˙ + T (t − s − h)x(s) ˙ − T (t − s)x(s). ˙ h

10.1. The abstract inhomogeneous Cauchy problem

125

The uniform boundedness of (T (t))t≥0 on any compact interval, the differentiability of x and the strong continuity of (T (t))t≥0 implies that the above expression converges to zero. Thus lim T (t − s − h)

h→0

x(s + h) − x(s) = T (t − s)x(s). ˙ h

Thus the function s → T (t − s)x(s) is differentiable on [0, t), and d (T (t − s)x(s)) = −AT (t − s)x(s) + T (t − s)(Ax(s) + f (s)) = T (t − s)f (s), ds where we used the equation x(s) ˙ = Ax(s)+f (s). Thus a classical solution to (10.2) necessarily has the form (10.3). Equation (10.3) is reminiscent of the variation of constants formula for ordinary differential equations, see (2.21). If x is a classical solution, then x is necessarily given by (10.3). Next we show that under suitable conditions (10.3) is already the (unique) classical solution of (10.2). Theorem 10.1.3. If A is the infinitesimal generator of a C0 -semigroup (T (t))t≥0 on a Hilbert space X, f ∈ C 1 ([0, τ ]; X) and x0 ∈ D(A), then the function x : [0, τ ] → X defined by (10.3) is continuously differentiable on [0, τ ] and it is the unique classical solution of (10.2). Proof. Uniqueness: If x1 and x2 are two different classical solutions of (10.2), then their difference Δ(t) = x1 (t) − x2 (t) satisfies the differential equation dΔ = AΔ, dt

Δ(0) = 0

and so we need to show that its only solution is Δ(t) ≡ 0. However, this follows directly from Lemma 5.3.2. t Existence: Clearly, all we need to show now is that t → v(t) = 0 T (t − s)f (s)ds 1 is an element of C ([0, τ ]; X), v(t) ∈ D(A) for every t ∈ [0, τ ] and v satisfies the differential equation (10.2) with x0 = 0. Now t s v(t) = f˙(α)dα ds T (t − s) f (0) + 0

0

t

t

t

T (t − s)f˙(α)dsdα,

T (t − s)f (0)ds +

= 0

0

(10.4)

α

where we have used Fubini’s Theorem. From Theorem 5.2.2.5, it follows that the t first term on the right hand side lies in the domain of A and that α T (t − s) f˙(α)ds ∈ D(A) for all α ∈ [0, t]. Furthermore, Theorem 5.2.2.5 implies t A T (t − s)f˙(α)ds = T (t − α)f˙(α) − f˙(α). (10.5) α

126


Since A is a closed operator, for every g ∈ C([0, τ ], X) with values in D(A) and Ag ∈ C([0, τ ], X), we have Ag(s) ds = A g(s) ds. Combining this fact with (10.4) and (10.5), we obtain that v(t) ∈ D(A), and t Av(t) = (T (t) − I)f (0) + (T (t − α) − I)f˙(α)dα 0 t = T (t)f (0) + T (t − α)f˙(α)dα − f (t). 0

Using the fact that the convolution product is commutative, i.e., t g(s)h(t − s)ds, we obtain 0

t 0

g(t−s)h(s)ds =

t

T (α)f˙(t − α)dα − f (t).

Av(t) = T (t)f (0) + 0

Moreover, we have t t d T (t)f (0) + T (α)f˙(t − α)dα − f (t) = T (α)f (t − α)dα − f (t) dt 0 0 = v(t) ˙ − f (t), and thus we may conclude that v(t) ˙ = Av(t) + f (t).

In the above proof, we have proved the following result. Corollary 10.1.4. If A is the infinitesimal generator of a C0 -semigroup (T (t))t≥0 on the Hilbert space X, and f ∈ C 1 ([0, τ ]; X), then the function v : [0, τ ] → X t defined by v(t) = 0 T (t− s)f (s)ds, t ∈ [0, τ ] is continuously differentiable on [0, τ ] and has values in the domain of A. Furthermore, it satisfies t T (t − α)f˙(α)dα − f (t). (10.6) Av(t) = T (t)f (0) + 0

The assumptions of Theorem 10.1.3 are too strong for control applications, where in general we do not wish to assume that f ∈ C 1 ([0, τ ]; X). Therefore we introduce the following weaker concept of a solution of (10.2). Definition 10.1.5. If f ∈ L1 ([0, τ ]; X) and x0 ∈ X, then we call the function x : [0, τ ] → X defined by (10.3) a mild solution of (10.2) on [0, τ ]. We note that (10.3) is a well-defined integral in the sense of Bochner or Pettis, (see Lemma A.1.6 and Example A.1.13). Of course, if f ∈ Lp ([0, τ ]; X) for some p ≥ 1, then necessarily f ∈ L1 ([0, τ ]; X). In applications, we usually find f ∈ L2 ([0, τ ]; X). Lemma 10.1.6. Assume that f ∈ L1 ([0, τ ]; X) and x0 ∈ X. The mild solution x defined by (10.3) is continuous on [0, τ ].


127

Proof. Since T (·)x0 is continuous, we can assume without loss of generality that x0 = 0. For δ > 0, consider

t

x(t + δ) − x(t) =

t+δ

(T (t + δ − s) − T (t − s))f (s)ds + 0

T (t + δ − s)f (s)ds t

t+δ

= (T (δ) − I)x(t) +

T (t + δ − s)f (s)ds. t

Then we estimate

t+δ

x(t + δ) − x(t) ≤ (T (δ) − I)x(t) + sup T (α)

f (s) ds

α∈[0,δ]

t

and the right-hand side converges to 0 as δ ↓ 0 by the strong continuity of the semigroup and Theorem 5.1.5.1. Now consider

t−δ

x(t − δ) − x(t) =

t

(T (t − δ − s) − T (t − s))f (s)ds −

T (t − s)f (s)ds, t−δ

0

noting that (T (t − δ − s) − T (t − s))f (s) is integrable, since f ∈ L1 ([0, τ ]; X) and using the properties of (T (t))t≥0 from Theorem 5.1.5.1 and Example A.1.13. An estimation of the integral above yields

t−δ

t

(T (t − δ − s) − T (t − s))f (s) ds +

x(t − δ) − x(t) ≤

T (t − s)f (s) ds. t−δ

0

Now (T (t − δ − s) − T (t − s))f (s) → 0 as δ ↓ 0, and by Theorem 5.1.5 there exists a constant Mt > 0, depending only on t, such that (T (t − δ − s) − T (t − s))f (s) ≤ Mt f (s) . So the first term converges to zero δ ↓ 0 by the Lebesgue Dominated Convergence Theorem, and the second term also tends to zero by similar arguments. In fact, the concept of a mild solution is the same as the concept of a weak solution used in the study of partial differential equations. Definition 10.1.7. Let f ∈ L1 ([0, τ ]; X). We call the function x : [0, τ ] → X a weak solution of (10.2) on [0, τ ] if the following holds: 1. x is continuous on [0, τ ]; 2. For all g ∈ C 1 ([0, τ ]; X) with g(t) ∈ D(A∗ ), t ∈ [0, τ ] and A∗ g ∈ L1 ([0, τ ]; X) we have τ τ ∗ g(t)+A ˙ g(t), x(t) dt+ g(t), f (t) dt = g(t), x(τ ) −g(0), x0 . (10.7) 0

0

128


We call x a weak solution of (10.2) on [0, ∞) if it is a weak solution on [0, τ ] for every τ ≥ 0. Theorem 10.1.8. For every x0 ∈ X and every f ∈ L1 ([0, τ ]; X) there exists a unique weak solution of (10.2), which is given by the mild solution of (10.2). Proof. Let x0 ∈ X and f ∈ L1 ([0, τ ]; X) be chosen arbitrarily and let x be the corresponding mild solution given by (10.3). Lemma 10.1.6 shows that x is continuous on [0, τ ]. Let g ∈ C 1 ([0, τ ]; X) with g(t) ∈ D(A∗ ), t ∈ [0, τ ] and A∗ g ∈ L1 ([0, τ ]; X). It is sufficient to prove that the function g(·), x(·) is absolutely continuous and its derivative is given by dg(t), x(t) = A∗ g(t), x(t) + g(t), f (t) dt

(10.8)

for almost all t. It is easy to show that g(·), x(·) is absolutely continuous. For h > 0 we have that g(t + h), x(t + h) − g(t), x(t) = g(t + h) − g(t), x(t + h) + g(t), x(t + h) − x(t) .

(10.9)

Since x is continuous and g is differentiable it is easy to see that g(t + h) − g(t), x(t + h) = g(t), ˙ x(t) . h→0 h lim

(10.10)

Furthermore, we have that x(t + h) − x(t) = x(t + h) − T (h)x(t) + T (h)x(t) − x(t) t+h T (t + h − s)f (s)ds + (T (h) − I)x(t). = t

Let λ be an element of the resolvent set of A. By the previous equation, Proposition A.1.7, and Theorem 5.2.2 we have for almost all t ∈ [0, τ ], (λI − A)−1 lim h↓0

x(t + h) − x(t) 1 t+h =(λI − A)−1 lim T (t + h − s)f (s)ds h↓0 h t h (λI − A)−1 (T (h) − I)x(t) + lim h↓0 h (T (h) − I)(λI − A)−1 x(t) = (λI − A)−1 f (t) + lim h↓0 h −1 = (λI − A) f (t) + A(λI − A)−1 x(t). (10.11)


129

Combining (10.9), (10.10), and (10.11) we find that lim h↓0

g(t + h), x(t + h) − g(t), x(t) = g(t), ˙ x(t) + limg(t), x(t + h) − x(t) h↓0 h ∗ −1 = g(t), ˙ x(t) + lim(λI − A) g(t), (λI − A) (x(t + h) − x(t)) h↓0

= g(t), ˙ x(t) + (λI − A)∗ g(t), (λI − A)−1 f (t) + A(λI − A)−1 x(t) = g(t), ˙ x(t) + g(t), f (t) + A∗ g(t), x(t) . This proves (10.8) for the derivative from the right. The proof for the derivative from the left goes similarly, see also the proof of Lemma 10.1.2. To prove the uniqueness, we use the fact that A∗ is the infinitesimal generator of the C0 -semigroup (T ∗ (t))t≥0 , see e.g. [10, 15, 24, 61]. If there are two weak solutions, then the difference xd (t) satisfies τ g(t) ˙ + A∗ g(t), xd (t) dt = g(τ ), xd (τ ) (10.12) 0

for all g ∈ C 1 ([0, τ ], X) with g(t) ∈ D(A∗ ), t ∈ [0, τ ] and A∗ g ∈ L1 ([0, τ ]; X). Take h ∈ C 1 ([0, τ ], X), and define z as the unique classical solution of z(t) ˙ = A∗ z(t) + h(t),

z(0) = 0.

By Lemma 10.1.2 this solution exists. We define g(t) = z(τ − t), t ∈ [0, τ ]. Using Lemma 10.1.2, it is easy to see that g ∈ C 1 ([0, τ ], X) and g(t) ∈ D(A∗ ) for all t ∈ [0, τ ]. Furthermore, g(t) ˙ = −z(τ ˙ − t) = −A∗ z(τ − t) − h(τ − t),

g(τ ) = 0.

Substituting this into equation (10.12), we obtain τ − h(τ − t), xd (t) dt = 0. 0

As this hold for all continuous functions h, and since xd is continuous, we conclude that xd is the zero function. In the following chapters, whenever we refer to the solution of the abstract evolution equation (10.2), we mean the mild solution (10.3). Example 10.1.9. In this example, we shall again consider the heat equation of Example 5.1.1. The model of the heated bar was given by ∂x ∂2x (ζ, t) = (ζ, t) + u(ζ, t), x(ζ, 0) = x0 (ζ), ∂t ∂ζ 2 ∂x ∂x (0, t) = 0 = (1, t). ∂ζ ∂ζ

130


We showed in Section 5.3 that if u = 0, then this partial differential equation can be formulated as an abstract differential equation on X = L2 (0, 1) of the form x(t) ˙ = Ax(t), t ≥ 0,

x(0) = x0 ,

where d2 h with dζ 2 + dh D(A) = h ∈ L2 (0, 1) | h, are absolutely continuous, dζ , d2 h dh dh 2 ∈ L (0, 1) and (0) = 0 = (1) . dζ 2 dζ dζ Ah =

(10.13)

(10.14)

We can include the control term in this formulation as follows: x(t) ˙ = Ax(t) + u(t), t ≥ 0,

x(0) = x0 ,

provided that u ∈ L1 ([0, τ ]; L2 (0, 1)). The solution is given by (10.3), which can be written as, see Example 5.1.4, x(ζ, t) =

∞

eλn t x0 , φn φn (ζ) +

n=0

1

=

z0 (α)dα + 0

t

∞

2e−n

n=1 1

u(α, s)dα +

+ 0

0

2

π2 t

eλn (t−s) u(·, s), φn (·) φn (ζ)ds

0 n=0

t ∞

t ∞

1

z0 (α) cos(nπα)dα cos(nπζ) 0

e−n

2

π 2 (t−s)

0 n=1

since λn = −n2 π 2 , n ≥ 0, φn (ζ) =

2

1

u(y, s) cos(nπα)dα cos(nπζ)ds, 0

√

2 cos(nπζ), n ≥ 1 and φ0 (ζ) = 1.

10.2 Outputs Usually, as in Chapter 2, the function f in the inhomogeneous abstract differential equation (10.2) is of the form Bu(·), where B is a bounded linear operator from the input space U to the state space X. We assume that U is also a Hilbert space. Thus the equation (10.2) reads x(t) ˙ = Ax(t) + Bu(t),

x(0) = x0 .

(10.15)

Again we assume that A is the infinitesimal generator of the C0 -semigroup (T (t))t≥0 . The state is x and the input is given by u. We denote the system (10.15) by Σ(A, B).

10.2. Outputs

131

In many systems the state is measured via an output, and so we add an output to the system Σ(A, B). We start by assuming that the output equation can be represented via a bounded operator. For the more general case, we refer to section 11.2. The system Σ(A, B) with an output equation is given by x(t) ˙ = Ax(t) + Bu(t), y(t) = Cx(t) + Du(t),

x(0) = x0 ,

(10.16) (10.17)

where A generates a C0 -semigroup (T (t))t≥0 on the state space X, B is a linear bounded operator from the input space U to X, C is a linear bounded operator from X to the output space Y , and D is a linear operator from U to Y . All spaces U , X and Y are supposed to be Hilbert spaces. We denote the system (10.16)–(10.17) by Σ(A, B, C, D). In Section 10.1 we showed that the mild solution of (10.16) is given by

t

T (t − s)Bu(s) ds.

x(t) = T (t)x0 + 0

This function is well-defined for every x0 ∈ X and every u ∈ L1 ([0, τ ]; U ), τ > 0. As the operators C and D are bounded, there is no difficulty in “solving” equation (10.17). We summarize the answer in the following theorem. Theorem 10.2.1. Consider the abstract equation (10.16)–(10.17), with A the infinitesimal generator of the C0 -semigroup (T (t))t≥0 , and B, C, and D bounded linear operators. The mild solution of (10.16)–(10.17) is given by the variation of constant formula (10.3)

t

T (t − s)Bu(s) ds, t T (t − s)Bu(s) ds + Du(t) y(t) = CT (t)x0 + C

x(t) = T (t)x0 +

0

0

for every x0 ∈ X and every u ∈ L1 ([0, τ ]; U ). As an example we return to Example 10.1.9 to which we add a measurement. Example 10.2.2. Consider the heated bar of Example 10.1.9 on which we measure the average temperature in the first half of the bar. ∂2x ∂x (ζ, t) = (ζ, t) + u(ζ, t), x(ζ, 0) = x0 (ζ), ∂t ∂ζ 2 ∂x ∂x (0, t) = 0 = (1, t), ∂ζ ∂ζ 1 y(t) = 02 x(ζ, t) dζ.

132


From Example 10.1.9 we have that the infinitesimal generator is given by (10.13) and (10.14). Furthermore, it is easy to see that the input operator equals the identity, i.e., U = X and B = I. The output space equals C and the output operator C is given by 12 Cx = x(ζ) dζ. (10.18) 0

This is clearly a bounded linear operator, and so this partial differential equation can be written in the form (10.16)–(10.17).

10.3 Bounded perturbations of C0 -semigroups In applications to control problems, the inhomogeneous term f in (10.2) is often determined by a control input of feedback type, namely, f (t) = Dx(t), where D ∈ L(X). This leads to the new Cauchy problem x(t) ˙ = (A + D)x(t), t ≥ 0,

x(0) = x0 ,

(10.19)

or in its integrated form

t

T (t − s)Dx(s)ds.

x(t) = T (t)x0 +

(10.20)

0

We expect that the perturbed system operator, A+D, is the infinitesimal generator of another C0 -semigroup (TD (t))t≥0 , such that the solution of (10.19) is given by x(t) = TD (t)x0 . This result is proved in Exercise 10.1 for the case that A generates a contraction semigroup. For the general case, we refer to [10, Section 3.2] or [15, page 158]. Theorem 10.3.1. Suppose that A is the infinitesimal generator of a C0 -semigroup (T (t))t≥0 on a Hilbert space X and that D ∈ L(X). Then A+D is the infinitesimal generator of a C0 -semigroup (TD (t))t≥0 on X. Moreover, if T (t) ≤ M eωt , then

TD (t) ≤ M e(ω+M D )t and for every x0 ∈ X the following equations are satisfied: t TD (t)x0 = T (t)x0 + T (t − s)DTD (s)x0 ds

(10.21)

(10.22)

0

and

t

TD (t − s)DT (s)x0 ds.

TD (t)x0 = T (t)x0 + 0

(10.23)

10.4. Exponential stabilizability

133

10.4 Exponential stabilizability In Chapter 4 we studied the stabilizability of the finite-dimensional system x(t) ˙ = Ax(t) + Bu(t),

x(0) = x0 .

In Sections 10.1 and 10.2 we saw that this equation is well-defined provided A is the infinitesimal generator of a C0 -semigroup and B is a bounded operator. Thus the concept of stabilizability generalizes naturally to an infinite-dimensional setting. Definition 10.4.1. Suppose that A is the infinitesimal generator of the C0 -semigroup (T (t))t≥0 on the Hilbert space X and that B ∈ L(U, X), where U is a Hilbert space. If there exists an F ∈ L(X, U ) such that A + BF generates an exponentially stable C0 -semigroup, (TBF (t))t≥0 , then we say that Σ(A, B) is exponentially stabilizable. An operator F ∈ L(X, U ) will be called a feedback operator. Remark 10.4.2. This definition differs from our definition of exponential stabilizability for finite-dimensional systems, see Definition 4.1.3. By Theorem 4.3.1, for finite-dimensional systems Definitions 4.1.3 and 10.4.1 are equivalent. For infinitedimensional systems this no longer holds, see also the Notes and References. As for finite-dimensional state spaces, we aim to characterize operators A and Bs such that Σ(A, B) is stabilizable. Unfortunately, such a characterization is not known. However, if the input space is finite-dimensional, or a little bit more general if the range of B is finite-dimensional, then a complete characterization of all systems Σ(A, B) which are exponentially stabilizable can be given. In this section we present this result. Hence we assume that B ∈ L(Cm , X), and that A is the infinitesimal generator of the C0 -semigroup (T (t))t≥0 on the Hilbert space X. For the proof of the main theorem we need the following lemma. Note that the order ν0 of the isolated eigenvalue λ0 is defined as follows. For every x ∈ X, limλ→λ0 (λ − λ0 )ν0 (λI − A)−1 x exists, but there exists an x0 ∈ X such that limλ→λ0 (λ − λ0 )ν0 −1 (λI − A)−1 x0 does not exist, see also page 103. For the eigenvalue λ0 of the operator Q with order ν0 < ∞ the multiplicity is defined as dim ker ((λ0 I − Q)ν0 ). Lemma 10.4.3. Consider the system Σ(A, B) with B ∈ L(Cm , X) and let F ∈ L(X, Cm ). For s ∈ ρ(A + BF ) we have the following properties: 1. −1 ∈ σ(BF (sI − A − BF )−1 ) if and only if −1 ∈ σ(F (sI − A − BF )−1 B) if and only if det(I + F (sI − A − BF )−1 B) = 0. If −1 ∈ σ(BF (sI − A − BF )−1 ), then −1 lies in the point spectrum. The same assertion holds for F (sI − A − BF )−1 B. 2. The order and the multiplicity of the eigenvalue −1 of BF (sI − A − BF )−1 and F (sI − A − BF )−1 B are finite and equal.

134


3. s ∈ ρ(A) if and only if −1 ∈ ρ(F (sI − A − BF )−1 B). If s ∈ ρ(A), then (sI − A)−1 satisfies (sI − A)−1 = (sI − A − BF )−1 − (sI − A − BF )−1 B −1 · I + F (sI − A − BF )−1 B F (sI − A − BF )−1 . (10.24) 4. If the holomorphic function s → det(I +F (sI −A−BF )−1 B), s ∈ ρ(A+BF ), is zero for s = s0 , but not identically zero in a neighbourhood of s0 , then s0 ∈ σ(A) and it is an eigenvalue of A with finite order and finite multiplicity. Proof. 1. The proof is based on the following equalities; and BF (sI − A − BF )−1 B = B F (sI − A − BF )−1 B F (sI − A − BF )−1 BF (sI − A − BF )−1 = F (sI − A − BF )−1 B F (sI − A − BF )−1 .

(10.25)

(10.26)

Since F (sI −A−BF )−1 B is an m×m-matrix all its spectral points are eigenvalues. Furthermore, since the range of the bounded linear operator BF (sI − A − BF )−1 is contained in Cm , it is a compact operator, and so every non-zero point in the spectrum is an isolated eigenvalue. This proves the second part of the first assertion. Let −1 be an eigenvalue of F (sI − A − BF )−1 B and let v ∈ Cm be the corresponding eigenvector, i.e, F (sI − A − BF )−1 Bv = −v. Thus in particular Bv = 0, and by multiplying this equation with B from the left and using (10.25) it follows that Bv is an eigenvector of BF (sI − A − BF )−1 with eigenvalue −1. The other implication can be proved in a similar manner. 2. From the previous part it follows that the dimension of the kernel of I + BF (sI − A − BF )−1 equals the dimension of the kernel of I + F (sI − A − ν BF )−1 B. Similarly the dimensions of the kernel of I + BF (sI − A − BF )−1 ν and I + F (sI − A − BF )−1 B are the same. Furthermore, since F (sI − A − BF )−1 B is an m × m and BF (sI − A − BF )−1 B is a compact operator the eigenvalue −1 is for both operators of finite order. Combining these two facts, we conclude that they have equal (finite) order and multiplicity. 3. We have the following simple equality: sI − A = I + BF (sI − A − BF )−1 (sI − A − BF ).

(10.27)

Since sI − A − BF is boundedly invertible, the operator sI − A is injective if and only if I + BF (sI − A − BF )−1 is injective. Using the same argument, we obtain that sI − A is surjective if and only if I + BF (sI − A − BF )−1 is surjective. Using the fact that BF (sI −A−BF )−1 is a compact operator, I +BF (sI −A−BF )−1 is


135

injective if and only if it is surjective if and only if −1 ∈ ρ(BF (sI −A−BF )−1 ). In particular, this implies that s ∈ ρ(A) if and only if −1 ∈ ρ(F (sI − A − BF )−1 B). Hence we have proved the first part. It is easy to see that the following equalities hold for s ∈ ρ(A + BF ): I = (sI − A)(sI − A − BF )−1 − BF (sI − A − BF )−1 , −1

B = (sI − A)(sI − A − BF )

−1

B − BF (sI − A − BF )

(10.28) B,

(10.29)

and hence B I + F (sI − A − BF )−1 B = (sI − A)(sI − A − BF )−1 B.

(10.30)

Thus for s ∈ ρ(A) ∩ ρ(A + BF ) we have −1 . (sI − A)−1 B = (sI − A − BF )−1 B I + F (sI − A − BF )−1 B

(10.31)

Multiplying (10.28) from the left by (sI − A)−1 and using (10.31) we find that (sI − A)−1 = (sI − A − BF )−1 − (sI − A)−1 BF (sI − A − BF )−1 = (sI − A − BF )−1 − (sI − A − BF )−1 B −1 · I + F (sI − A − BF )−1 B F (sI − A − BF )−1 . Thus we have shown (10.24). 4. Since det(I + F (sI − A − BF )−1 B) is not identically zero in a neighbourhood of s0 , part 1 implies that s0 is an isolated eigenvalue of A. It remains to show that the order and multiplicity of s0 ∈ σ(A) are finite. Let ν0 be the order of s0 as a zero of det(I + F (sI − A − BF )−1 B). Using (10.24) we find that lim (s − s0 )ν0 (sI − A)−1 x = lim (s − s0 )ν0 (sI − A − BF )−1 x

s→s0

s→s0

− lim (s − s0 ) (sI − A − BF )−1 B s→s0 −1 F (sI − A − BF )−1 x. · I + F (sI − A − BF )−1 B ν0

By the definition of ν0 and the fact that (sI − A − BF )−1 is holomorphic on a neighbourhood of s0 , the limit on the right hand side exists. Hence the order of s0 as an eigenvalue of A cannot be larger than ν0 . That the multiplicity is finite follows from (10.27), see also part 1. Lemma 10.4.3 implies that if s ∈ ρ(A + BF ) ∩ σ(A), then s is an eigenvalue of A. In fact, we show that if Σ(A, B) is stabilizable then A can have at most finitely many eigenvalues in C+ 0 . Such a separation of the spectrum is an important property of the generator. We introduce the following notation: + σ + := σ(A) ∩ C+ 0 ; C0 = {λ ∈ C | Re(λ) > 0}, −

σ := σ(A) ∩

C− 0;

C− 0

= {λ ∈ C | Re(λ) < 0}.

(10.32) (10.33)

136


Definition 10.4.4. A satisfies the spectrum decomposition assumption at zero if σ + is bounded and separated from σ − in such a way that a rectifiable, simple, closed curve, Γ, can be drawn so as to enclose an open set containing σ + in its interior and σ − in its exterior. From Theorem 8.2.4 follows that if the spectral decomposition assumption at zero holds, then we have a corresponding decomposition of the state space X and of the operator, A. More precisely, the spectral projection PΓ defined by 1 P Γx = (λI − A)−1 xdλ, (10.34) 2πi Γ where Γ is traversed once in the positive direction (counterclockwise), induces the following decomposition: X = X + ⊕ X −,

where X + := P Γ X and X − := (I − P Γ )X.

In view of this decomposition, it is convenient to use the notation + + + A T (t) B 0 0 A= , T (t) = , B = , 0 A− 0 T − (t) B−

(10.35)

(10.36)

where B + = P Γ B ∈ L(U, X + ), and B − = (I − P Γ )B ∈ L(U, X − ). In fact, we have decomposed our system Σ(A, B) as the vector sum of the two subsystems: Σ(A+ , B + ) on X + and Σ(A− , B − ) on X − . The following theorem shows that if the system Σ(A, B) is stabilizable, then X + is finite-dimensional, and (T − (t))t≥0 is exponentially stable. Theorem 10.4.5. Consider the system Σ(A, B) on the state space X and input space Cm . The following assertions are equivalent: 1. Σ(A, B) is exponentially stabilizable; 2. Σ(A, B) satisfies the spectrum decomposition assumption at zero, X + is finite-dimensional, (T − (t))t≥0 is exponentially stable, and the finite-dimensional system Σ(A+ , B + ) is controllable, where we have used the notation introduced in equations (10.35) and (10.36). If Σ(A, B) is exponentially stabilizable, then a stabilizing feedback operator is given by F = F + P Γ , where F + is a stabilizing feedback operator for Σ(A+ , B + ). Proof. 2 ⇒ 1. Since the finite-dimensional system Σ(A+ , B + ) is controllable, there exists a feedback operator F + ∈ L(X + , Cm ) such that the spectrum of A+ +B + F + + m lies in C− 0 . Choose the feedback operator F = [F 0] ∈ L(X, C ) for the system Σ(A, B). The perturbed operator A + BF =

A+ +B + F + 0 B− F + A− + ∈ C0 we have

generates a C0 -

semigroup by Theorem 10.3.1. Furthermore, for s that, see Exercise 10.2, (sI − A+ − B + F + )−1 0 −1 (sI − A − BF ) = . (sI − A− )−1 B − F + (sI − A+ − B + F + )−1 (sI − A− )−1


137

Since the semigroups generated by A+ + B + F + and A− are exponentially stable, Theorem 8.1.4 implies that the corresponding resolvent operators are uniformly bounded in the right half-plane. Thus (sI − A − BF )−1 is uniformly bounded in C+ 0 , and thus by Theorem 8.1.4 its corresponding semigroup is exponentially stable. 1 ⇒ 2. By assumption there exists an operator F ∈ L(X, Cm ) such that A + BF generates an exponentially stable C0 -semigroup. By Definition 8.1.1, there exist constants M > 0 and γ < 0, such that

TBF (t) ≤ M eγt .

(10.37)

From Lemma 10.4.3, for every s ∈ {s ∈ C | Re(s) > γ} we have s ∈ σ(A) if and only if det(I + F (sI − A − BF )−1 B) = 0. Now the determinant det(I + F (·I − A − BF )−1 B) is holomorphic on {s ∈ C | Re(s) > γ} and therefore there cannot be an accumulation point of zeros in {s ∈ C | Re(s) ≥ γ + ε}, ε > 0, unless the determinant is identically zero. From (10.37), it follows that for all ε > 0,

∞

e2(−γ−ε)t F TBF (t)B 2 dt < ∞

0

and by the Paley-Wiener Theorem A.2.9, we deduce that F ((· + γ + ε)I − A − BF )−1 B ∈ H2 (Cm×m ).

(10.38)

By Lemma A.2.6 this implies that lim

ρ→∞

sup s∈C+ γ+ε ,|s|≥ρ

F (sI − A − BF )−1 B = 0.

Consequently, det(I + F (·I − A − BF )−1 B) cannot be identically zero in C+ 0 , and the function has no finite accumulation point there. Moreover, we can always find a sufficiently large ρ such that

F (·I − A − BF )−1 B ≤

1 in C+ 0 \D(ρ), 2

(10.39)

−1 where D(ρ) = {s ∈ C+ B is invertible 0 | |s| ≤ ρ}. Thus I + F (sI − A − BF ) + for all s ∈ C0 \ D(ρ). Inside the compact set D(ρ) a holomorphic function has at most finitely many zeros, and applying Lemma 10.4.3 we see that σ + comprises at most finitely many points with finite multiplicity. Hence the spectrum decomposition assumption holds at zero. From Theorem 8.2.4.4 and e follows that X + = ran P + is finite-dimensional and σ(A+ ) = σ + ⊂ C+ 0 . Thus it remains to show that (T − (t))t≥0 is exponentially stable and that Σ(A+ , B + ) is controllable.

138

Chapter 10. Inhomogeneous Abstract Differential Equations By Lemma 10.4.3 we have that (sI − A)−1 = (sI − A − BF )−1 − (sI − A − BF )−1 B −1 · I + F (sI − A − BF )−1 B F (sI − A − BF )−1 .

(10.40)

Since A + BF generates an exponentially stable semigroup, by Theorem 8.1.4 we obtain that (·I − A − BF )−1 is uniformly bounded in C+ 0 . Combining this with (10.39), equation (10.40) implies that sup

(sI − A)−1 < ∞.

s∈C+ 0 \D(ρ)

From Lemma 8.2.6 we conclude that (T − (t))t≥0 is exponentially stable. Finally, we prove that the system Σ(A+ , B + ) is controllable. By Theorem 10.3.1 we obtain that t P Γ T (t − s)BF TBF (s)x0 ds. P Γ TBF (t)x0 = P Γ T (t)x0 + 0

For x0 ∈ X + this equation reads

t

T + (t − s)B + F TBF (s)x0 ds.

P Γ TBF (t)x0 = T + (t)x0 + 0

Since (TBF (t))t≥0 is exponentially stable, for any x0 ∈ X + there exists an input u ∈ L1loc ([0, ∞); Cm ), i.e., u(t) = F TBF (t)x0 , such that the solution of the finitedimensional system x(t) ˙ = A+ x(t) + B + u(t),

x(0) = x0

is exponentially decaying. By Definition 4.1.3, the system Σ(A+ , B + ) is stabilizable. Since all eigenvalues of the matrix A+ lie in C+ 0 , using Theorem 4.3.3 we may conclude that Σ(A+ , B + ) is controllable. We apply the above result to the system of Example 10.1.9. Example 10.4.6. Here we consider again the heat equation of Example 10.1.9, but we choose a different input operator. ∂x ∂2x (ζ, t) = (ζ, t) + ½[ 12 ,1] (ζ)u(t), ∂t ∂ζ 2 ∂x ∂x (0, t) = 0 = (1, t). ∂ζ ∂ζ

x(ζ, 0) = x0 (ζ),

(10.41) (10.42)

As in Example 10.1.9 we can write this partial differential equation as an abstract differential equation on X = L2 (0, 1): x(t) ˙ = Ax(t) + Bu(t), t ≥ 0,

x(0) = x0 ,

10.5. Exercises

139

where A is given by (10.13) and (10.14) and B ∈ L(C, X) is given by (Bu) (ζ) = ½[ 12 ,1] (ζ)u. Since σ(A) ∩ C+ 0 = {0}, the system is not exponentially stable. Using the fact that X possesses an orthonormal basis of eigenvectors of A, see Example 5.1.4, it is easy to see that P Γ x = x, φ0 φ0 and thus X + = span{φ0 }, X − = spann≥1 {φn }. Furthermore, the semigroup (T − (t))t≥0 is given by T − (t)x =

∞

e−n

2

π2 t

x, φn φn

n=1

and therefore the semigroup (T − (t))t≥0 is exponentially stable. If we can show that Σ(A+ , B + ) is controllable, then Theorem 10.4.5 implies that the system (10.41)– (10.42) is exponentially stabilizable. We have that A+ = 0 and B + = P Γ B = ½[ 12 ,1] , 1 1 = 12 . Using Theorem 3.1.6 it is easy to see that the one-dimensional system Σ(0, 12 ) is controllable, and therefore the controlled heat equation is exponentially stabilizable.

10.5 Exercises 10.1. In this exercise we prove Theorem 10.3.1 in the case that A generates a contraction semigroup. Hence throughout this exercise we assume that A is the infinitesimal generator of the contraction semigroup (T (t))t≥0 on the Hilbert space X. (a) Show that the operator A + D − 2 D I with domain D(A) generates a contraction semigroup on X. Use Exercise 5.3 to conclude that A + D with domain D(A) generates a C0 -semigroup on X. (b) Denote by (TD (t))t≥0 the C0 -semigroup generated by A + D. Let x0 ∈ D(A). Use Theorem 10.1.3 and show that x(t) := TD (t)x0 − t T (t)x0 − 0 T (t − s)DTD (s)x0 ds is a classical solution of x(t) ˙ = Ax(t),

x(0) = 0.

Use Lemma 5.3.2 to conclude that x(t) = 0 for t ≥ 0. (c) Let x0 ∈ D(A). Show that x(t) = T (t)x0 + classical solution of x(t) ˙ = (A + D)x(t),

t 0

TD (t − s)DT (s)x0 ds is a

x(0) = x0 .

140

Chapter 10. Inhomogeneous Abstract Differential Equations (d) Prove that (10.22) and (10.23) hold.

10.2. Let A1 and A2 be two closed, densely defined operators on X1 and X2 , respectively. Let Q be a bounded operator from X1 to X2 . Show that if A1 and A2 are (boundedly) invertible, then the following operator A1 0 , D(Aext ) = D(A1 ) ⊕ D(A2 ), Aext = Q A2 is invertible as well, and the inverse is given by A−1 0 1 A−1 = . −1 ext −A−1 A−1 2 QA1 2 10.3. Consider the partial differential equation ∂x ∂2x (ζ, t) = (ζ, t) + 10x(ζ, t) + ½[0, 14 ] (ζ)u(t), ∂t ∂ζ 2 ∂x ∂x (0, t) = 0 = (1, t) ∂ζ ∂ζ 1 y(t) = x(ζ, t) dζ − u(t).

x(ζ, 0) = x0 (ζ),

(10.43) (10.44) (10.45)

0

(a) Formulate the partial differential equation (10.43)–(10.45) as a system of the form (10.16)–(10.17) on the state X = L2 (0, 1). (b) Is the system exponentially stabilizable?

10.6 Notes and references The results as formulated in Section 10.1 and 10.3 are well-documented in the literature and can be found in any standard text on C0 -semigroups, such as [24], [61], and [15]. However, since these results are of eminent importance for applications, textbooks aimed on applications of semigroup theory contain these results as well, see [10] or [44]. Since the output equation is important for control theory only, the results of Section 10.2 are not treated in books on C0 -semigroups. For more results on this, and on the state space system Σ(A, B, C, D) we refer to [10]. In our definition of exponential stabilizability, Definition 10.4.1, we have chosen the control as u(t) = F x(t). For the finite-dimensional case we could equivalently allow for locally L1 -functions. This equivalence no longer holds for infinitedimensional systems, see Example 2.13 of [48]. However, if the (stabilizing) input u is chosen as an L2 ([0, ∞); U )-function, then there exists a bounded F such that A + BF generates an exponentially stable semigroup. The proof uses optimal control theory, see e.g. Chapter 6 of [10].


141

The implication 2 to 1 in Theorem 10.4.5 has been known since the mid 1970s of the last century. However, the reserve implication stayed open for at least ten years. Then it was proved independently of each other by Desch and Schappacher [11], Jacobson and Nett [27], and Nefedov and Sholokhovich [42].

Chapter 11

Boundary Control Systems In this chapter we are in particular interested in systems with a control at the boundary of their spatial domain. We show that these systems have well-defined solutions provided the input is sufficiently smooth. The simplest boundary control system is most likely the controlled transport equation, which is given by ∂x ∂x (ζ, t) = (ζ, t), ∂t ∂ζ x(ζ, 0) = x0 (ζ), x(1, t) = u(t),

ζ ∈ [0, 1], t ≥ 0 ζ ∈ [0, 1] t ≥ 0,

(11.1)

where u denotes the control function. In Exercise 6.3 we have solved this partial differential equation for the specific choice u = 0. In this chapter we show that the solution of (11.1) is given by " x0 (ζ + t), ζ + t ≤ 1, x(ζ, t) = u(ζ + t − 1), ζ + t > 1.

11.1 Boundary control systems Boundary control problems like (11.1) occur frequently in applications, but unfortunately they do not fit into our standard formulation x(t) ˙ = Ax(t) + Bu(t),

x(0) = x0 .

(11.2)

However, for sufficiently smooth inputs it is possible to reformulate these problems in such a way that they lead to an associated system in the standard form (11.2). First we explain the idea behind this reformulation for the system (11.1). Assume that x is a classical solution of the p.d.e. (11.1) and that u is continuously differentiable. Defining v(ζ, t) = x(ζ, t) − u(t), B. Jacob and H.J. Zwart, Linear Port-Hamiltonian Systems on Infinite-dimensional Spaces, Operator Theory: Advances and Applications 223, DOI 10.1007/978-3-0348-0399-1_11, © Springer Basel 2012

143

144

Chapter 11. Boundary Control Systems

we obtain the following partial differential equation for v ∂v ∂v (ζ, t) = (ζ, t) − u(t), ˙ ∂t ∂ζ v(1, t) = 0, t ≥ 0.

ζ ∈ [0, 1], t ≥ 0

This partial differential equation for v can be written in the standard form as v(t) ˙ = Av(t) + B u ˜(t) for u˜ = u. ˙ Hence via a simple trick, we can reformulate this p.d.e. with boundary control into a p.d.e. with internal control. The price we have to pay is that u has to be smooth. The trick applied to (11.1) can be extended to abstract control systems of the form x(t) ˙ = Ax(t), x(0) = x0 , (11.3) Bx(t) = u(t), where A : D(A) ⊂ X → X is linear, the control function u takes values in the Hilbert space U , and the boundary operator B : D(B) ⊂ X → U is linear and satisfies D(A) ⊂ D(B). In order to reformulate equation (11.3) into an abstract form (11.2), we need to impose extra conditions on the system. Definition 11.1.1. The control system (11.3) is a boundary control system if the following hold: 1. The operator A : D(A) → X with D(A) = D(A) ∩ ker(B) and Ax = Ax

for x ∈ D(A)

(11.4)

is the infinitesimal generator of a C0 -semigroup (T (t))t≥0 on X; 2. There exists an operator B ∈ L(U, X) such that for all u ∈ U we have Bu ∈ D(A), AB ∈ L(U, X) and BBu = u,

u ∈ U.

(11.5)

Part 2 of the definition implies in particular that the range of the operator B equals U . Note that part 1 of the definition guarantees that the system (11.3) possesses a unique solution for the choice u = 0, i.e., the homogeneous equation is well-posed. Part 2 allows us to choose every value in U for the input u(t) at time t ≥ 0. In other words, the values of an input are not restricted, which is a plausible condition. We say that the function x : [0, τ ] → X is a classical solution of the boundary control system of Definition 11.1.1 on [0, τ ] if x is a continuously differentiable function, x(t) ∈ D(A) for all t ∈ [0, τ ], and x(t) satisfies (11.3) for all t ∈ [0, τ ].

11.1. Boundary control systems

145

The function x : [0, ∞) → X is a classical solution on [0, ∞) if x is a classical solution on [0, τ ] for every τ > 0. For a boundary control system, we can apply a similar trick as the one applied to the transport equation. This is the subject of the following theorem. It turns out that v(t) = x(t) − Bu(t) is the solution of the abstract differential equation v(t) ˙ = Av(t) − B u(t) ˙ + ABu(t),

v(0) = v0 .

(11.6)

Since A is the infinitesimal generator of a C0 -semigroup and B and AB are bounded linear operators, Theorem 10.1.3 implies that equation (11.6) has a unique classical solution for v0 ∈ D(A) and u ∈ C 2 ([0, τ ]; U ). Furthermore, we can prove the following relation between the (classical) solutions of (11.3) and (11.6). Theorem 11.1.2. Consider the boundary control system (11.3) and the abstract differential equation (11.6). Assume that u ∈ C 2 ([0, τ ]; U ). If v0 = x0 − Bu(0) ∈ D(A), then the classical solutions of (11.3) and (11.6) on [0, τ ] are related by v(t) = x(t) − Bu(t).

(11.7)

Furthermore, the classical solution of (11.3) is unique. Proof. Suppose that v is a classical solution of (11.6). Then for t ∈ [0, τ ] we get v(t) ∈ D(A) ⊂ D(A) ⊂ D(B), Bu(t) ∈ D(B), and Bx(t) = B(v(t) + Bu(t)) = Bv(t) + BBu(t) = u(t), where we have used that v(t) ∈ D(A) ⊂ ker B and that equation (11.5) holds. Furthermore, equation (11.7) implies x(t) ˙ = v(t) ˙ + B u(t) ˙ = Av(t) − B u(t) ˙ + ABu(t) + B u(t) ˙

by (11.6)

= Av(t) + ABu(t) = A(v(t) + Bu(t))

by (11.4)

= Ax(t)

by (11.7).

Thus, if v is a classical solution of (11.6), then x defined by (11.7) is a classical solution of (11.3). The other implication is proved similarly. The uniqueness of the classical solutions of (11.3) follows from the uniqueness of the classical solutions of (11.6), see Theorem 10.1.3. Remark 11.1.3. In Theorem 11.1.2 we assume that v0 = x0 − Bu(0) ∈ D(A). However, this can equivalently be formulated in a condition on x0 and u(0). More precisely, the following equivalence is easy to prove, see Exercise 11.1: If (11.3) defines a boundary control system, then the difference x0 − Bu(0) is an element of the domain of A if and only if x0 ∈ D(A) and Bx0 = u(0).

146

Chapter 11. Boundary Control Systems The (mild) solution of (11.6) is given by

t

T (t − s) (ABu(s) − B u(s)) ˙ ds

v(t) = T (t)v0 +

(11.8)

0

for every v0 ∈ X and every u ∈ H 1 ([0, τ ]; U ), τ > 0. Therefore, the function

t

x(t) = T (t)(x0 − Bu(0)) +

T (t − s) (ABu(s) − B u(s)) ˙ ds + Bu(t)

(11.9)

0

is called the mild solution of the abstract boundary control system (11.3) for every x0 ∈ X and every u ∈ H 1 ([0, τ ]; U ) τ > 0. As an example we study again the controlled transport equation of equation (11.1). Example 11.1.4. We consider the system ∂x ∂x (ζ, t) = (ζ, t), ∂t ∂ζ x(ζ, 0) = x0 (ζ),

ζ ∈ [0, 1], t ≥ 0 ζ ∈ [0, 1] t≥0

x(1, t) = u(t),

for an input u ∈ H 1 (0, τ ). In order to write this example in the form (11.3) we choose X = L2 (0, 1) and dx , dζ Bx = x(1), Ax =

D(A) = H 1 (0, 1), D(B) = D(A).

These two operators satisfy the assumption of a boundary control system. More precisely: the operators A and B are linear, A restricted to the domain D(A)∩ker B generates a C0 -semigroup, see Exercise 6.3. Furthermore, the range of B is C = U and the choice B = ½[0,1] implies BBu = u. Using the fact that AB = 0, we conclude from equation (11.9) that the mild solution is given by

t

x(t) = T (t)(x0 − Bu(0)) + = T (t)(x0 − u(0)) −

T (t − s)(ABu(s) − B u(s)) ˙ ds + Bu(t) 0 t

T (t − s)½[0,1] u(s) ˙ ds + u(t).

0

Using the precise representation of the shift-semigroup, see Exercise 6.3, we can write the solution of the boundary controlled partial differential equation as x(ζ, t) = (x0 (ζ + t) − u(0))½[0,1] (ζ + t) −

0

t

½[0,1] (ζ + t − s)u(s) ˙ ds + u(t).

11.2. Outputs for boundary control systems

147

If ζ + t > 1, we have x(ζ, t) = −u(τ )|tζ+t−1 + u(t) = u(ζ + t − 1), and if ζ + t ≤ 1, then x(ζ, t) = x0 (ζ + t) − u(0) − u(τ )|t0 + u(t) = x0 (ζ + t). Or equivalently,

" x(ζ, t) =

x0 (ζ + t), ζ + t ≤ 1, u(ζ + t − 1), ζ + t > 1,

(11.10)

which proves our claim made in the introduction of this chapter. It is easy to show that this example cannot be written as an abstract control system x(t) ˙ = Ax(t) + Bu(t), (11.11) with B a bounded operator. If the controlled transport equation were of the form (11.11), then Theorem 10.1.3 would imply that x(t) ∈ D(A), whenever x(0) ∈ D(A) and u ∈ C 1 ([0, τ ]; X). Choosing x0 = 0, t = 12 , and u = 1, (11.10) implies that x(ζ, 12 ) = 0 if ζ ≤ 12 and x(ζ, 12 ) = 1 if ζ > 12 , which is in contradiction to x( 12 ) ∈ D(A) ⊂ D(A) = H 1 (0, 1). Summarizing, the boundary controlled transport equation of this example cannot be written in the form (11.11). The controlled transport equation is a simple example of our general class of port-Hamiltonian systems. This example could be written as a boundary control system. In section 11.3 we show that this holds in general for a port-Hamiltonian system. However, before we do this, we add an output to the boundary control system.

11.2 Outputs for boundary control systems In the previous section we showed that a system with control at the boundary possesses a (unique) classical solution, provided the input and initial condition are smooth enough. In Section 10.2, we saw that having a solution for the state equation, easily led to the solution of the output equation. In this section, we show that for classical solutions of a boundary control system a similar property holds. We add a (boundary) output to the boundary control system (11.3). x(t) ˙ = Ax(t), Bx(t) = u(t), Cx(t) = y(t)

x(0) = x0 ,

(11.12) (11.13) (11.14)

where (A, B) satisfies the conditions of a boundary control system, see Definition 11.1.1 and C is a linear operator from D(A) to Y with Y a Hilbert space.

148


Since a classical solution of (11.12)–(11.13) takes values in the domain of A, and since C is well-defined on this space, there is no difficulty in solving (11.14). We summarize the answer in the following theorem. Theorem 11.2.1. Consider the boundary control system (11.12)–(11.14) with (A, B) satisfying the conditions of Definition 11.1.1 and C a linear operator from D(A) to Y . If u ∈ C 2 ([0, τ ]; U ), x0 ∈ D(A), and Bx0 = u(0), then the classical solution of (11.12)–(11.14) is given by

t

x(t) = T (t)(x0 − Bu(0)) +

T (t − s)(ABu(s) − B u(s)) ˙ ds + Bu(t), t y(t) = CT (t)(x0 − Bu(0)) + C T (t − s)(ABu(s) − B u(s)) ˙ ds + CBu(t). 0

0

If D(A) is a Hilbert space, and C is a bounded linear operator from D(A) to Y , then the output y is a continuous function. For an example of a boundary control system with a boundary output, we refer to Example 11.3.6. This example is a port-Hamiltonian system with boundary control and observation. In the following section we show that these systems fall into the class of boundary control systems.

11.3 Port-Hamiltonian systems as boundary control systems In this section we add a boundary control to a port-Hamiltonian system and we show that the assumptions of a boundary control system are satisfied. The portHamiltonian system with control is given by ∂ ∂x (ζ, t) = P1 (H(ζ)x(ζ, t)) + P0 (H(ζ)x(ζ, t)), ∂t ∂ζ f∂ (t), u(t) = WB,1 e∂ (t) f∂ (t) 0 = WB,2 . e∂ (t)

(11.15) (11.16) (11.17)

We make the following assumptions. Assumption 11.3.1. • P1 ∈ Kn×n is invertible and self-adjoint; • H ∈ L∞ ([a, b]; Kn×n ), H(ζ) is self-adjoint for a.e. ζ ∈ [a, b] and there exist M, m > 0 such that mI ≤ H(ζ) ≤ M I for a.e. ζ ∈ [a, b];

11.3. Port-Hamiltonian systems as boundary control systems • WB :=

WB,1 WB,2

149

∈ Kn×2n has full rank.

Thus, in particular, P1 and H satisfy the assumptions of Definition 7.1.2. We recall that the boundary effort and boundary flow are given by, see (7.26) f∂ (Hx)(b) = R0 , e∂ (Hx)(a) where R0 is the invertible 2n × 2n-matrix defined in (7.27). We can write the port-Hamiltonian system (11.15)–(11.17) as a boundary control system x(t) ˙ = Ax(t), x(0) = x0 , Bx(t) = u(t), by defining ∂ Ax = P1 (Hx) + P0 (Hx), ∂ζ + , f∂ 2 n 1 n D(A) = x ∈ L ([a, b]; K ) | Hx ∈ H ([a, b]; K ), WB,2 =0 , e∂ f Bx = WB,1 ∂ , e∂ D(B) = D(A).

(11.18) (11.19) (11.20) (11.21)

Following Chapter 7, we choose the Hilbert space X = L2 ([a, b]; Kn ) equipped with the inner product 1 b f, g X := f (ζ)∗ H(ζ)g(ζ) dζ (11.22) 2 a as the state space. The input space U equals Km , where m is the number of rows of WB,1 1 . We are now in the position to show that the controlled port-Hamiltonian system is indeed a boundary control system. Theorem 11.3.2. If the operator Ax = P1

∂ (Hx) + P0 (Hx) ∂ζ

with the domain + , f∂ WB,1 1 n D(A) = x ∈ X | Hx ∈ H ([a, b]; K ), ∈ ker e∂ WB,2

(11.23)

(11.24)

generates a C0 -semigroup on X, then the system (11.15)–(11.17) is a boundary control system on X. 1 Note that m has two meanings: It is used as a lower-bound for H and as the dimension of our input space.

150


Proof. Equations (11.19) and (11.20) imply that D(A) = D(A) ∩ ker B, and hence part 1 of Definition 11.1.1 is satisfied. The n × 2n-matrix WB has full rank n and R0 is an invertible matrix. Thus there exists a 2n × n-matrix S such that WB,1 I 0 WB R0 S = R0 S = m , (11.25) 0 0 WB,2 m where Im is the identity matrix choice for the matrix S is

on K . A possible S11 S12

−1 ∗ ∗ −1 Im 0 . We write S = , where S11 and S21 are S = R0 WB (WB WB ) S21 S22 0 0 n × m-matrices, and we define the operator B ∈ L(Km , X) by ζ −a b−ζ (Bu)(ζ) := H(ζ)−1 S11 + S21 u. b−a b−a

The definition of B implies that Bu is a square integrable function and that

= HBu ∈ H 1 ([a, b]; Kn ). Furthermore, from (11.25) it follows that WB,2 R0 SS11 21 0. Combining this with the definition of the boundary effort and boundary flow, we obtain that Bu ∈ D(A). Furthermore, B and AB are linear bounded operators from Km to X and using (11.25) once more, we obtain S BBu = WB,1 R0 11 u = u. S21

Thus the port-Hamiltonian system is indeed a boundary control system.

Remark 11.3.3. An essential condition in the above theorem is that A given by (11.23) with domain (11.24) generates a C0 -semigroup. Theorem 7.2.4 and Assumption 11.3.1 imply that this holds in particular when P0∗ = −P0 and WB [ 0I I0 ] WB∗ ≥ 0. Since the term P0 H can be seen as a bounded perturbation of (11.18) with P0 = 0, Theorem 7.2.4 and Theorem 10.3.1 show that A given by (11.23) with domain (11.24) generates a C0 -semigroup when WB [ 0I I0 ] WB∗ ≥ 0. As an example we once more study the controlled transport equation. Example 11.3.4. We consider the system ∂x ∂x (ζ, t) = (ζ, t), ∂t ∂ζ x(ζ, 0) = x0 (ζ),

ζ ∈ [0, 1], t ≥ 0

(11.26)

ζ ∈ [0, 1].

This system can be written as a port-Hamiltonian system (11.15) by choosing n = 1, P0 = 0, P1 = 1 and H = 1. Therefore, we have 1 1 −1 1 x(1, t) − x(0, t) f∂ (t) R0 = √ = √ and . e∂ (t) 2 1 1 2 x(1, t) + x(0, t)

11.3. Port-Hamiltonian systems as boundary control systems

151

Since n = 1, we have the choice of either applying one control or no control at all. Adding a boundary control, the control can be written as, see (11.16),

1 x(1, t) − x(0, t) 1 u(t) = a b √ = √ [(a + b)x(1, t) + (b − a)x(0, t)] . 2 x(1, t) + x(0, t) 2 (11.27) Note that WB = [a b] has full rank if and only if a2 + b2 = 0. We assume this from now on. By Theorem 11.3.2 we only have to check whether the homogeneous partial differential equation generates a C0 -semigroup. Using Remark 11.3.3 this holds if 2ab ≥ 0. Thus possible boundary controls for the transport equation are √ 2 ), u(t) = x(1, t), (a = b = √ 2 √ u(t) = 3x(1, t) − x(0, t), (a = 2, b = 2 2). However, this remark does not provide an answer for the control u(t) = −x(1, t) + 3x(0, t). By using other techniques it can be shown that (11.26) with this control is also a boundary control system, see Chapter 13. Next we focus on a boundary observation for port-Hamiltonian systems. We develop conditions on the boundary observation guaranteeing that a certain balance equation is satisfied, which is important in Chapters 12 and 13. The standard Hamiltonian system with boundary control and boundary observation is given by ∂ (Hx(t)) + P0 (Hx(t)), ∂ζ f∂ (t) u(t) = WB , e∂ (t) f (t) y(t) = WC ∂ . e∂ (t) x(t) ˙ = P1

(11.28) (11.29) (11.30)

It is assumed that P1 , H and WB satisfy the conditions of Assumption 11.3.1. Note that we have taken WB = WB,1 or equivalently WB,2 = 0. In other words, we are using the maximal number of controls. The output equation is formulated very similar to the control equation. So we assume that the output space Y = Kk , and thus WC is a matrix of size k × 2n. Since we want the outputs to be independent, we assume that WC has full rank. Furthermore, since we do not want to measure quantities that B we already have chosen as an input, see (11.29), we assume that the matrix W WC has full rank. If we have full measurements, i.e., k = n, then the above assumptions imply B

that the matrix W with the fact that WC is invertible. Combining this assumption B

∗ ∗ WB WC Σ = [ 0I I0 ] is invertible, we obtain that the product W Σ [ ] is invertible WC as well. Its inverse is defined as −1

−1 ∗ WB WB ΣWB∗ WB ΣWC∗ ∗ PWB ,WC := Σ WB WC = . (11.31) WC WC ΣWB∗ WC ΣWC∗

152


We choose the same state space as in Section 11.3, i.e, X = L2 ([a, b]; Kn ) with inner product (11.22). Theorem 11.3.5. Consider (11.28)–(11.30), satisfying Assumption Bthe

system (k+n)×2n having full rank. 11.3.1, WC ∈ Kk×2n and W WC ∈ K Assume that the operator A defined by (11.23) and (11.24) generates a C0 semigroup on for every u ∈ C 2 ([0, ∞); Kn ), Hx(0) ∈ H 1 ([a, b]; Kn ), and X. Then f∂ (0) u(0) = WB e∂ (0) , the system (11.28)–(11.30) has a unique (classical) solution,

with Hx(t) ∈ H 1 ([a, b]; Kn ), t ≥ 0, and the output y is continuous. Furthermore, if additionally P0∗ = −P0 and k = n, then the following balance equation is satisfied for every t ≥ 0:

d 1 ∗ u(t) 2 ∗ u (t) y (t) PWB ,WC

x(t) X = . (11.32) y(t) dt 2 Proof. Theorem 11.3.2 implies that the system (11.28)–(11.29) is a boundary control system on X, where the operators A and B are given by ∂ f Bx = WB ∂ , Ax = P1 (Hx) + P0 (Hx), e∂ ∂ζ 3 4 2 n 1 D(A) = D(B) = x ∈ L ([a, b]; K ) | Hx ∈ H ([a, b]; Kn ) .

Let A and B be the corresponding operators satisfying the properties of Definition 11.1.1. By assumption we have u ∈ C 2 ([0, ∞); Kn ), x(0) ∈ D(A), and Bx(0) = u(0). Thus Remark 11.1.3 and Theorem 11.1.2 imply that (11.28)–(11.29) possesses a unique classical solution with Hx(t) ∈ H 1 ([a, b]; Kn ), t ≥ 0. Combining equations (11.30) and (11.14), we see that (Hx)(b) Cx = WC R0 . (Hx)(a) This is clearly a linear bounded operator from the domain of A to Ck , and thus Theorem 11.2.1 implies that the output corresponding to a classical solution is a continuous function. If P0 = −P0∗ and k = n, then we may apply Theorem 7.1.5. Using this theorem and the fact that Σ = [ I0 I0 ], the classical solution x satisfies d 1

x(t) 2X = (f∂∗ (t)e∂ (t) + e∗∂ (t)f∂ (t)) dt 2

1 ∗ f∂ (t) ∗ f (t) e∂ (t) Σ = e∂ (t) 2 ∂ 1 ∗ u (t) 2 1 ∗ u (t) = 2 =

which completes the proof.

−1

−1 WB u(t) WC∗ Σ WC y(t)

u(t) y ∗ (t) PWB ,WC , y(t)

y ∗ (t) WB∗

11.3. Port-Hamiltonian systems as boundary control systems

153

As an example we consider the vibrating string. Example 11.3.6. We consider the vibrating string on the spatial interval [a, b] for u2

u1

Figure 11.1: The vibrating string with two controls which the forces at both ends are the control variables, and the velocities at these ends are the outputs. In Example 7.1.1 we saw that the model is given by ∂2w 1 ∂ ∂w T (ζ) (ζ, t) , (11.33) (ζ, t) = ∂t2 ρ(ζ) ∂ζ ∂ζ where ζ ∈ [a, b] is the spatial variable, w(ζ, t) is the vertical position of the string at position ζ and time t, T is the Young’s modulus of the string, and ρ is the mass density. This system has the energy/Hamiltonian 2 2 ∂w ∂w (ζ, t) + T (ζ) (ζ, t) dζ. ρ(ζ) (11.34) ∂t ∂ζ a ρ ∂w ∂t , P1 = [ 01 10 ], Furthermore, it is a port-Hamiltonian system with state x = ∂w ∂ζ 1 0 P0 = 0, and H(ζ) = ρ(ζ) . The boundary effort and boundary flow for the 0 T (ζ) vibrating string are given by, see Example 7.2.5, ∂w ∂w ∂w 1 T (b) ∂w 1 ∂ζ (b) − T (a) ∂ζ (a) ∂t (b) + ∂t (a) √ , e = f∂ = √ . ∂ ∂w ∂w ∂w ∂w 2 2 T (b) ∂ζ (b) + T (a) ∂ζ (a) ∂t (b) − ∂t (a) (11.35) As depicted in Figure 11.1, the forces at the ends are the inputs and the velocity at both ends are the outputs, i.e., 5 6 5 6 ∂w T (b) ∂w (b, t) ∂ζ (b, t) ∂t u(t) = , y(t) = ∂w . (11.36) T (a) ∂w ∂ζ (a, t) ∂t (a, t) 1 E(t) = 2

b

Using (11.35) we find that WB and WC are given as 1 1 1 0 0 1 0 1 1 0 WB = √ , WC = √ . (11.37) 2 −1 0 0 1 2 0 −1 1 0 B

WB and WC are 2 × 4-matrices of full rank and W WC is invertible. Furthermore, 0 0 WB ΣWB∗ = . 0 0

154


By Theorem 7.2.4 we conclude that the associated A generates a contraction semigroup on the energy space. In fact, it generates a unitary group, see Exercise 7.2. Hence our system satisfies the conditions of Theorem 11.3.5. In particular, it is a boundary control system, and since P0 = 0 the balance equation (11.32) holds. Since 0 0 1 0 , and WC ΣWB∗ = , WC ΣWC∗ = 0 0 0 −1 we find that d 1

x(t) 2X = (u∗1 (t)y1 (t) − u∗2 (t)y2 (t) + y1∗ (t)u1 (t) − y2∗ (t)u2 (t)) . dt 2

(11.38)

Note that this equals equation (7.3).

11.4 Exercises 11.1. Prove the assertion in Remark 11.1.3 11.2. Consider the transmission line of Exercise 7.1 for any of the following boundary condition: (a) At the left-end the voltage equal u1 (t) and at the right-end the voltage equals the input u2 (t). (b) At the left-end we put the voltage equal to zero and at the right-end the voltage equals the input u(t). (c) At the left-end we put the voltage equal to u(t) and at the right-end the voltage equals R times the current, for some R > 0. Show that these systems can be written as a boundary control system. 11.3. Consider the vibrating string of Example 7.1.1 with the boundary conditions ∂w (a, t) = 0 ∂t

and

T (b)

∂w (b, t) = u(t) t ≥ 0. ∂ζ

(a) Reformulate this system as a boundary control system. (b) Prove that we have a well-defined input-output system, if we measure the velocity at the right-hand side, y(t) = ∂w ∂t (b, t). (c) Does a balance equation like (11.32) hold? 11.4. In the formulation of port-Hamiltonian systems as boundary control systems, we have the possibility that some boundary conditions are set to zero, see (11.17). However, when we add an output, this possibility was excluded, see (11.28)–(11.30). In this exercise we show that this did not pose a restriction to the theory.


155

(a) Show that if WB,2 = 0, i.e., WB = WB,1 , and WB ΣWB∗ ≥ 0, then (11.15) with control (11.16) is a well-defined boundary control system. What is the domain of the infinitesimal generator A? (b) Let WB be a n × 2n matrix of full rank satisfying WB ΣWB∗ ≥ 0. We

f∂ decompose u = WB e∂ as WB,1 f∂ u1 = . u= u2 WB,2 e∂ Show that the choice u2 = 0 is allowed. Furthermore, show that it leads to the same boundary control system as in (11.15)–(11.17). 11.5. Consider the coupled strings of Exercise 7.4. Now we apply a force u(t), to the bar in the middle, see Figure 11.2. This implies that the force balance in

11 00 00 11 00 11 00 11 00 11

1 0 0 1 0 1 0 1 0 1

u(t)

I

II

III

11 00 00 11 00 11 00 11 00 11 00 11 00 11 00 11

Figure 11.2: Coupled vibrating strings with external force the middle reads TI (b)

∂wI ∂wII ∂wIII (b) = TII (a) (a) + TIII (a) (a) + u(t). ∂ζ ∂ζ ∂ζ

(a) Formulate the coupled vibrating strings with external force as a boundary control system. (b) Additionally, we measure the velocity of the bar in the middle. Reformulate the system with this choice for the output as a system of form (11.28)–(11.30). (c) For the input and output defined above, determine the power balance in terms of the input and output, see (11.32).

11.5 Notes and references The boundary control system as presented in this chapter was introduced by Fattorini [16], and has become a standard way of reformulating partial differential equations with boundary control as an abstract system. The application to portHamiltonian systems of Sections 11.3 and 11.2 can be found in [36].

Chapter 12

Transfer Functions In this chapter we introduce the concept of transfer functions. In the system and control literature a transfer function is usually defined via the Laplace transform. However, in this chapter we use a different approach. Nevertheless, we explain the Laplace transform-approach by means of an example, that is, we consider the ordinary differential equation y¨(t) + 3y(t) ˙ − 7y(t) = −u(t) ˙ + 2u(t),

(12.1)

where the dot denotes the derivative with respect to time. Let L denote the Laplace transform, and let F be the Laplace transform of the function f , i.e., (L(f )) (s) = F (s). Recall that the following rules hold for the Laplace transform L(f˙) (s) = sF (s) − f (0), L(f¨) (s) = s2 F (s) − sf (0) − f˙(0). Assuming that y(0) = y(0) ˙ = u(0) = 0 and applying the Laplace transform to the differential equation (12.1), we obtain the algebraic equation s2 Y (s) + 3sY (s) − 7Y (s) = −sU (s) + 2U (s).

(12.2)

Thus we get the following relation between the Laplace transform of u and y: Y (s) =

−s + 2 U (s). s2 + 3s − 7

(12.3)

The rational function s2−s+2 +3s−7 is called the transfer function associated to the differential equation (12.1). This is the standard approach to transfer functions. However, this approach faces some difficulties, especially when we want to extend the concept to partial B. Jacob and H.J. Zwart, Linear Port-Hamiltonian Systems on Infinite-dimensional Spaces, Operator Theory: Advances and Applications 223, DOI 10.1007/978-3-0348-0399-1_12, © Springer Basel 2012

157

158

Chapter 12. Transfer Functions

differential equations. One of the difficulties is that the functions u and y have to be Laplace transformable. Considering u as input or control and y as output, the assumption that u is Laplace transformable is not very restrictive. However, once u is chosen, y is given by the differential equation, and it is a priori not known whether y is Laplace transformable. Another difficulty is that the Laplace transform of a function only exists in some right half-plane of the complex plane, which implies that equality (12.3) only holds for those s in the right-half plane for which the Laplace transform of u and y both exist. The right-half plane in which the Laplace transform of a function exists is named the region of convergence. Even for the simple differential equation (12.1) equality (12.3) does not hold everywhere. Taking into account the region of convergence of both u and y, equation (12.3) can only hold for those s which lie right of the zeros of the polynomial s2 + 3s − 7. However, for applications it is important to know the transfer function on a larger domain in the complex plane. For the finite-dimensional system (12.1) the transfer for all s ∈ C, where s not a zero of s2 +3s−7. function G is given by G(s) = s2−s+2 +3s−7 To overcome all these difficulties and to justify G(s) = s2−s+2 +3s−7 on a larger domain in the complex plane, we define the transfer function in a different way. We try to find solutions of the differential equation which are given as exponential functions. Again we illustrate this approach for the simple differential equation (12.1). Given s ∈ C, we try to find a solution pair of the form (u, y) = (est , ys est )t≥0 . If for an s such a solution exists, and it is unique, then we call ys the transfer function of (12.1) in the point s. Substituting this solution pair into the differential equation, we obtain t ≥ 0. (12.4) s2 ys est + 3sys est − 7ys est = −sest + 2est , We recognize the common term est which is never zero, and hence (12.4) is equivalent to s2 ys + 3sys − 7ys = −s + 2. (12.5) This is uniquely solvable for ys if and only if s2 + 3s − 7 = 0. Furthermore, ys is given by ys = s2−s+2 +3s−7 . Hence it is possible to define the transfer function without running into mathematical difficulties. The same approach works well for p.d.e.’s and abstract differential equations as the concept of a solution is well-defined.

12.1 Basic definition and properties In this section we start with a very general definition of a transfer function, which even applies to systems not described by a p.d.e, but via e.g. a difference differential equation or an integral equation. Therefore, we first introduce the notion of a general system. Let T := [0, ∞) be the time axis. Furthermore, we distinguish three spaces, U , Y , and R. U and Y are the input- and output space, respectively, whereas R contains the remaining variables. In our examples, R will become the state space X. Since s ∈ C, the exponential solutions will be complex-valued.

12.1. Basic definition and properties

159

Therefore we assume that U, R and Y are complex Hilbert spaces. A system S is a subset of L1loc ([0, ∞); U × R × Y ), i.e., a subset of all locally integrable functions from the time axis T to U × R × Y . Note that two functions f and g are equal in L1loc ([0, ∞); U × R × Y ) if f (t) = g(t) for almost every t ≥ 0. Definition 12.1.1. Let S be a system, s be an element of C, and u0 ∈ U . We say that (u0 est , r(t), y(t))t≥0 is an exponential solution in S if there exist r0 ∈ R, y0 ∈ Y , such that (u0 est , r0 est , y0 est ) = (u0 est , r(t), y(t)) for a.e. t ≥ 0. Let s ∈ C. If for every u0 ∈ U there exists an exponential solution, and the corresponding output trajectory y0 est , t ∈ [0, ∞) is unique, then we call the mapping u0 → y0 the transfer function at s. We denote this mapping by G(s). Let Ω ⊂ C be the set consisting of all s for which the transfer function at s exists. The mapping s ∈ Ω → G(s) is defined as the transfer function of the system S. In our applications r will usually be the state x. We begin by showing that the transfer function for the system x(t) ˙ = Ax(t) + Bu(t), y(t) = Cx(t) + Du(t),

(12.6) (12.7)

see (10.16) and (10.17), where B, C and D are bounded operators and A generates a strongly continuous semigroup, exists and is given by the formula G(s) = C(sI − A)−1 B + D, for s in the resolvent set, ρ(A), of A. Theorem 12.1.2. Consider the linear system (12.6)–(12.7), with B, C, and D bounded operators. The solution of the system is given by the mild solution, see Theorem 10.2.1. If (u(t), x(t), y(t))t≥0 is an exponential solution of (12.6)–(12.7), then x is a classical solution of (12.6). Furthermore, for s ∈ ρ(A), the transfer function exists and is given by (12.8) G(s) = C(sI − A)−1 B + D. Proof. The mild solution of (12.6) with initial condition x(0) = x0 is uniquely determined and given by t T (t − τ )Bu(τ )dτ. (12.9) x(t) = T (t)x0 + 0

For an exponential solution this equation should equal t st T (t − τ )Bu0 esτ dτ. x0 e = T (t)x0 +

(12.10)

0

Taking x0 = (sI − A)−1 Bu0 , the right-hand side of this equation can be written as t T (t − τ )(sI − A)x0 esτ dτ T (t)x0 + 0 t st = T (t)x0 + e T (t − τ )e−s(t−τ ) (sI − A)x0 dτ. 0

160


By Exercise 5.3 the infinitesimal generator of the C0 -semigroup (T (t)e−st )t≥0 is given by A − sI. Applying Theorem 5.2.2.4 to the above equation we find that T (t)x0 + e

st 0

t

T (t − τ )e−s(t−τ ) (sI − A)x0 dτ = T (t)x0 − est T (t)e−st x0 − x0 = est x0 .

(12.11)

Thus by choosing x0 = (sI − A)−1 Bu0 , we obtain an exponential solution. We show next that this solution is unique. If there are two exponential solutions for the same u0 and s, then by (12.10) we see that their difference satisfies x0 , x ˜0 est = T (t)˜

t ≥ 0,

(12.12)

˜0 is an eigenvector of T (t) for all t ≥ 0. Since the left-hand for some x ˜0 ∈ X. Thus x side is differentiable so is the right-hand side. Thus x˜0 ∈ D(A) and A˜ x0 = s˜ x0 , see also Exercise 8.5. Since s ∈ ρ(A), this implies that x ˜0 = 0. Thus the exponential solution is unique. The output equation of the system yields y0 est = y(t) = Cx(t) + Du(t) = Cx0 est + Du0 est = C(sI − A)−1 Bu0 est + Du0 est . Thus for every s ∈ ρ(A) the output trajectory corresponding to an exponential solution is uniquely determined, and y0 = C(sI − A)−1 Bu0 + Du0 . This implies that the system possesses a transfer function on Ω = ρ(A) and the transfer function is given by (12.8). Theorem 12.1.2 shows that for linear systems of the form (12.6)–(12.7) the transfer function exists, and is given by G(s) = C(sI − A)−1 B + D. Unfortunately, port-Hamiltonian systems can in general not be written in the form (12.6)–(12.7) with bounded operators B, C and D. However, by Theorem 11.3.2 every portHamiltonian system is a boundary control system. In order to calculate the transfer function of port-Hamiltonian systems, we first focus on transfer functions of boundary control systems. Again we show that every exponential solution is a classical solution. Theorem 12.1.3. Consider the system x(t) ˙ = Ax(t), u(t) = Bx(t), y(t) = Cx(t),

x(0) = x0 , (12.13)

where (A, B) satisfies the conditions of a boundary control system, see Definition 11.1.1 and C is a linear operator from D(A) to Y , where Y is a Hilbert space. The solution of the state differential equation is given by the mild solution, see equation (11.9).

12.1. Basic definition and properties

161

Every exponential solution of (12.13) is also a classical solution. Furthermore, for s ∈ ρ(A), the transfer function exists and is given by G(s) = C(sI − A)−1 (AB − sB) + CB.

(12.14)

For s ∈ ρ(A) and u0 ∈ U , G(s)u0 can also be calculated as the (unique) solution of sx0 = Ax0 , u0 = Bx0 , G(s)u0 = Cx0 ,

(12.15)

with x0 ∈ D(A). Proof. By (11.9) the mild solution of (12.13) is given by

t

x(t) = T (t)(x0 − Bu(0)) +

T (t − τ ) (ABu(τ ) − B u(τ ˙ )) dτ + Bu(t). 0

Assuming that (u(t), x(t), y(t))t≥0 is an exponential solution, the above equation becomes t T (t − τ ) (ABesτ u0 − Bsesτ u0 ) dτ + Best u0 . (12.16) est x0 = T (t)(x0 − Bu0 ) + 0

Applying the Laplace transform, for λ ∈ C with Re(λ) > max{Re(s), ω0 }, where ω0 is the growth bound of the semigroup (T (t))t≥0 , we obtain su0 x0 u0 u0 = (λI − A)−1 (x0 − Bu0 ) + (λI − A)−1 AB −B , +B λ−s λ−s λ−s λ−s or equivalently, x0 − Bu0 u0 su0 −1 −1 = (λI − A) (x0 − Bu0 ) + (λI − A) −B AB . λ−s λ−s λ−s This implies that x0 − Bu0 ∈ D(A), and (λI − A)(x0 − Bu0 ) = (x0 − Bu0 )(λ − s) + ABu0 − Bsu0 . Subtracting the term λ(x0 − Bu0 ) from both sides, we obtain (sI − A)(x0 − Bu0 ) = ABu0 − Bsu0 ,

(12.17)

or equivalently, for s ∈ ρ(A), x0 is uniquely determined as x0 = (sI − A)−1 (AB − sB) u0 + Bu0 .

(12.18)

162


Hence this shows that if we have for s ∈ ρ(A) an exponential solution (u(t), x(t))t≥0 = (u0 est , x0 est )t≥0 , then the x0 is uniquely determined by u0 . Starting with x0 defined by (12.18) and reading backwards, we see from equation (12.16) that x(t) = x0 est is the solution for u(t) = u0 est . By (12.18) we have that x0 − Bu0 ∈ D(A) and since u ∈ C 2 ([0, ∞); U ), Theorem 11.1.2 implies that x is a classical solution of (12.13). In particular, for all t ≥ 0, x(t) ∈ D(A). By assumption the domain of C contains the domain of A. Hence y0 est = y(t) = Cx(t), holds point-wise in t. The choice t = 0 implies y0 = Cx0 . Combining this equality with (12.18), we obtain that the transfer function is well-defined and is given by (12.14). Since x(t) = x0 est is the classical solution corresponding to the input u(t) = st u0 e and the initial condition x(0) = x0 , the differential equation (12.13) reads sx0 est = Ax0 est , u0 est = Bx0 est , y0 est = Cx0 est , which is equivalent to equation (12.15). The uniqueness of y0 follows from the uniqueness of x0 , see (12.18). We close this section by calculating the transfer function of a controlled transport equation. Example 12.1.4. Consider the system ∂x ∂x (ζ, t) = (ζ, t), ∂t ∂ζ x(ζ, 0) = x0 (ζ), u(t) = x(1, t), y(t) = x(0, t),

ζ ∈ [0, 1], t ≥ 0,

(12.19)

ζ ∈ [0, 1], t ≥ 0, t ≥ 0.

If we define Cx = x(0), then it is easy to see that all assumptions of Theorem 12.1.3 are satisfied, see Theorem 11.3.5. Hence we can calculate the transfer function using equation (12.15). For the system (12.19), equation (12.15) reads dx0 (ζ), dζ u0 = x0 (1),

sx0 (ζ) =

G(s)u0 = x0 (0). The solution of this differential equation is given by x0 (ζ) = αesζ . Using the other two equations, we see that G(s) = e−s .

12.2. Transfer functions for port-Hamiltonian systems

163

12.2 Transfer functions for port-Hamiltonian systems In this section we apply the results of the previous section to the class of portHamiltonian systems. Due to the fact that every port-Hamiltonian system is a boundary control system provided the corresponding operator A generates a C0 -semigroup, it is a direct application of Theorem 12.1.3. Recall that a portHamiltonian system with boundary control and boundary observation is given by ∂x ∂ (ζ, t) = P1 (H(ζ)x(ζ, t)) + P0 (H(ζ)x(ζ, t)), ∂t ∂ζ f (t) u(t) = WB ∂ , e∂ (t) f (t) y(t) = WC ∂ , e∂ (t)

(12.20) (12.21) (12.22)

see (11.28)–(11.30). It is assumed that (12.20)–(12.21) satisfy the assumption of a port-Hamiltonian system, see Section 11.3. Further we assume that WB ΣWBT ≥ 0, where Σ = [ 0I I0 ]. This guarantees that (12.20)–(12.21) is a boundary control system, see Remark 11.3.3. We assume we have k (independent) measurements, i.e., WC is a full rank

B is of full rank. matrix of size k × 2n and W WC Theorem 12.2.1. The transfer function G(s) of the system (12.20)–(12.22) is uniquely determined by d (Hx0 ) + P0 (Hx0 ), dζ f∂,0 u0 = WB , e∂,0 f G(s)u0 = WC ∂,0 , e∂,0 sx0 = P1

where

f∂,0 (Hx0 ) (b) = R0 . e∂,0 (Hx0 ) (a)

(12.23) (12.24) (12.25)

(12.26)

This transfer function has the following properties: 1. If P0 , P1 , WB , and WC are real matrices and H is real-valued, then G(s) is a real matrix for real s. 2. If we have full measurements, i.e., k = n, then the transfer function satisfies the equality

1 ∗ u0 2 ∗ ∗ u u0 G(s) PWB ,WC Re (s) x0 = (12.27) G(s)u0 2 0 B

∗ W ∗ WC where PWB ,WC is the inverse of W ]. WC Σ [ B

164


Proof. The proof is a direct combination of Theorems 11.3.5 and 12.1.3. The first theorem implies that the system (12.20)–(12.22) is a well-defined boundary control system and that the output equation is well-defined in the domain of the system operator A. Hence all conditions of Theorem 12.1.3 are satisfied, and equation (12.15) for the port-Hamiltonian system reads (12.23)–(12.25). 1. To prove the assertion in part 1, we need some notation. We denote by v the element-wise complex conjugate of the vector v. Similarly, for the matrix Q. With this notation we find from equation (12.23) that sx0 = sx0 = P1

d d (Hx0 ) + P0 (Hx0 ) = P1 (Hx0 ) + P0 (Hx0 ), dζ dζ

where we have used that s, P1 , P0 and H are real-valued. Similarly, we find f u0 = WB ∂,0 , e∂,0 f G(s)u0 = WC ∂,0 , e∂,0 where

(Hx0 ) (b) f∂,0 = R0 . (Hx0 ) (a) e∂,0

Thus G(s) satisfies the equations (12.23)–(12.26) with u0 replaced u0 . However, u0 is arbitrary, and so we see that G(s) is also the transfer function at s. Since the transfer function is unique, we conclude that G(s) = G(s) and thus G is real-valued. 2. The transfer function is by definition related to the exponential solution (u0 est , x0 est ,

G(s)u0 est )t≥0 .

Hence if we have full measurement, then we may substitute this solution in (11.32). Dividing by e2Re(s)t gives (12.27). By (12.23)–(12.25), the calculation of the transfer function is equivalent to solving an ordinary differential equation. If H is constant, i.e., independent of ζ, this is relatively easy. However, in general it can be very hard to solve this ordinary differential equation analytically, see Exercise 12.2. In Theorem 12.2.1 we assumed that there are exactly n controls. However, this does in general not hold and the transfer function can also be calculated if some of the boundary conditions are set to zero. There are two possibilities to calculate the transfer function in this more general situation. Exercise 11.4 shows that this more general system satisfies all the conditions of Theorem 12.1.3, and hence Theorem 12.1.3 can be used to obtain the differential equation determining the transfer function. Another approach is to regard the zero boundary conditions

12.2. Transfer functions for port-Hamiltonian systems

165

as additional inputs, and to add extra measurements such that we have n controls and n measurements. The requested transfer function is now a sub-block of the n × n-transfer function. We explain this in more detail by means of an example, see Example 12.2.3. Note that (12.27) equals the balance equation (7.16). Example 12.2.2. Consider the following partial differential equation ∂hx ∂x (ζ, t) = (ζ, t), ∂t ∂ζ x(ζ, 0) = x0 (ζ), u(t) = h(b)x(b, t), y(t) = h(a)x(a, t),

ζ ∈ [a, b], t ≥ 0, ζ ∈ [a, b], t ≥ 0,

(12.28)

t ≥ 0,

where h : [a, b] → R is a (strictly) positive continuous function. It is easy to see that this is a port-Hamiltonian system with P1 = 1, P0 = 0, and H = h. The energy balance equation (7.16) becomes d b x(ζ, t)∗ h(ζ)x(ζ, t) dζ = |h(b)x(b, t)|2 − |h(a)x(a, t)|2 . (12.29) dt a We derive the transfer function by applying Theorem 12.2.1 to the partial differential equation (12.28). Thus we obtain the following ordinary differential equation dhx0 (ζ), dζ u0 = h(b)x0 (b),

sx0 (ζ) =

ζ ∈ [a, b], (12.30)

G(s)u0 = h(a)x0 (a). The solution of this differential equation is given by c0 exp(p(ζ)s), x0 (ζ) = h(ζ) where c0 is a constant not depending on ζ, and ζ p(ζ) = h(α)−1 dα. a

The second equation of (12.30) can be used to calculate c0 . Using the third equation of (12.30) we obtain (12.31) G(s) = e−p(b)s . The strict positivity of h on [a, b] implies that |G(s)| ≤ 1 for all s with non-negative real part. This result also follows from the balance equation (12.29) without calculating the transfer function as follows. Since every exponential solution is a classical solution, for the exponential solution (u0 est , x0 est , G(s)u0 est )t≥0 we obtain d b st 2 |e | x0 (ζ)∗ h(ζ)x0 (ζ) dζ = |u0 est |2 − |G(s)u0 est |2 . (12.32) dt a

166


Using the fact that |est |2 = e2Re(s)t , equation (12.29) implies that for all u0 and all s ∈ C with Re(s) ≥ 0 we have that |G(s)u0 |2 ≤ |u0 |2 . Since u0 is a scalar, we conclude that the absolute value of G(s) is bounded by 1 for Re(s) ≥ 0. Example 12.2.3. Consider the vibrating string of Example 11.3.6, as depicted in Figure 12.1. Again we assume that we control the forces at both ends, and meau2

u1

Figure 12.1: The vibrating string with two controls sure the velocities at the same points. However, now we assume that the Young’s modulus T and the mass density ρ are constant and that our spatial interval is [0, 1]. Hence the model becomes ∂2w 1 ∂ ∂w T (ζ, t) , ζ ∈ [0, 1], (12.33) (ζ, t) = ∂t2 ρ ∂ζ ∂ζ 6 5 T ∂w ∂ζ (1, t) u(t) = , (12.34) T ∂w ∂ζ (0, t) 5 6 ∂w (1, t) ∂t y(t) = ∂w . (12.35) ∂t (0, t) Here w(ζ, t) is the vertical position of thestring at position ζ and time t. We write this system using the state variable x = ∂x1 ∂ (ζ, t) = (T x2 (ζ, t)) , ∂t ∂ζ T x2 (1, t) , u(t) = T x2 (0, t) 1 x1 (1, t) y(t) = ρ1 . ρ x1 (0, t)

ρ ∂w ∂t ∂w ∂ζ

, and find

∂x2 ∂ (ζ, t) = ∂t ∂ζ

x1 (ζ, t) ρ

,

(12.36) (12.37) (12.38)

In order to calculate the transfer function we have to solve the ordinary differential equation dx20 1 dx10 (ζ), sx20 (ζ) = (ζ), sx10 (ζ) = T dζ ρ dζ T x20 (1) u u0 = 10 = , T x20 (0) u20 1 x10 (1) y y0 = 10 = ρ1 . y20 ρ x10 (0)

(12.39) (12.40) (12.41)

12.3. Exercises

167

It is easy to see that the solution of (12.39) is given by x10 (ζ) = αeλsζ + βe−λsζ ,

x20 (ζ) =

αλ λsζ βλ −λsζ e − e , ρ ρ

$ where λ = Tρ , and α, β are complex constants. Using equation (12.40) we can relate the constants α and β to u0 , λ α 1 −e−λs = λs u0 . β e − e−λs 1 −eλs

(12.42)

(12.43)

Combining this with (12.41), gives y0 =

λs λ e + e−λs λs −λs 2 ρ(e − e )

Thus the transfer function is given by 5 1 λ tanh(λs) G(s) = 1 ρ sinh(λs)

−2 u . −eλs − e−λs 0

6 1 − sinh(λs) . 1 − tanh(λs)

(12.44)

Using now the balance equation (11.38), we find 2Re (s) x0 2 = Re (u∗10 y10 ) − Re (u∗20 y20 ) = Re (u∗10 G11 (s)u10 + u∗10 G12 (s)u20 ) − Re (u∗20 G21 (s)u10 + u∗20 G22 (s)u20 ).

(12.45)

Choosing u20 = 0, we conclude that the real part of G11 (s) is positive for Re(s) > 0. An analytic function f on C+ 0 := {s ∈ C | Re (s) > 0} which satisfies Ref (s) ≥ 0 for Re(s) > 0 and f (s) ∈ R for s ∈ (0, ∞), is called positive real. Since physical parameters are real, we conclude from Theorem 12.2.1 that the transfer function is real-valued for real s. Using the fact that G11 is analytic on s ∈ C+ 0 , equation (12.45) implies that G11 is positive real. This can also be checked by direct λ calculation on G11 (s) = ρ tanh(λs) . Next we consider the system defined by the p.d.e. (12.33) with input u(t) = ∂w ∂w T ∂w ∂ζ (0, t), output y(t) = ∂t (1, t) and boundary condition T ∂ζ (1, t) = 0. We could proceed as we did above. However, we can easily obtain the transfer function by choosing u10 = 0 in (12.40) and only considering y10 in (12.41). Hence the transfer λ function of this system is − ρ sinh(λs) .

12.3 Exercises 12.1. Consider the vibrating string of Example 7.1.1. Thus in particular, we choose ∂w ∂t (a, t) = 0. We control this system by applying a force at the right-hand side u(t) = T (b) ∂w ∂ζ (b, t) and we measure the velocity at the same position, ∂w i.e., y(t) = ∂t (b, t).

168

Chapter 12. Transfer Functions (a) Show that the transfer function is positive real. Hint: See Exercise 11.3 (b) Determine the transfer function for the case that the functions ρ and T are constant. (c) Next choose T (ζ) = eζ , and ρ(ζ) = 1, and determine the transfer function for this case. Hint: You may use a computer package like Maple or Mathematica.

12.2. Consider the system of the transmission line of Exercise 7.1. As input we take the voltage at the right-hand side, and as output we take the current at the left-hand side. Furthermore, the voltage at the left-hand side is set to zero. Determine the transfer function under the assumption that the capacity C and the inductance L are constant. Hint: See also Exercise 11.2. 12.3. In this exercise we prove some well-known properties of transfer functions. Let S1 and S2 be two systems, i.e, S1 ⊂ L1loc ([0, ∞); U1 × R1 × Y1 ) and S2 ⊂ L1loc ([0, ∞); U2 ×R2 ×Y2 ). Assume that for a given s ∈ C both systems have a transfer function. Furthermore, we assume that for both systems the output is determined by the input and the state, that is, if (u0 est , r0 est , y(t))t≥0 ∈ S, then y(t) = y0 est for some y0 and for almost every t ≥ 0. (a) Show that for boundary control systems the output is determined by the input and the state, see Theorem 12.1.3. (b) Assume that Y1 = U2 . The series connection Sseries ⊂ L1loc ([0, ∞); U1 × (R1 × R2 ) × Y2 ) of S1 and S2 is defined as follows, see Figure 12.2, (u1 , (r1 , r2 ), y2 ) ∈ Sseries if there exists a y1 such that (u1 , r1 , y1 ) ∈ S1 and (y1 , r2 , y2 ) ∈ S2 . Show that the series connection Sseries has the transfer function G(s) = G2 (s)G1 (s) at s. u1

S1

y1

u2

S2

y2

Figure 12.2: Series connection (c) Assume that U1 = U2 and Y1 = Y2 . The parallel connection Sparallel ⊂ L1loc ([0, ∞); U1 × (R1 × R2 ) × Y2 ) of S1 and S2 is defined as follows: (u1 , (r1 , r2 ), y) ∈ Sparallel if there exists a y1 ∈ Y and y2 ∈ Y2 such that (u1 , r1 , y1 ) ∈ S1 , (u1 , r2 , y2 ) ∈ S2 , and y = y1 + y2 .


169

Show that the parallel connection Sparallel has the transfer function G1 (s) + G2 (s) at s. u1

S1

y1 y

u u2

S2

y2

Figure 12.3: Parallel connection (d) Assume that U1 = Y2 and Y1 = U2 . The feedback connection Sfeedback ⊂ L1loc ([0, ∞); U1 × (R1 × R2 ) × Y1 ) of S1 and S2 is defined as follows: (u, (r1 , r2 ), y1 ) ∈ Sfeedback if there exists a u1 and y2 such that (u1 , r1 , y1 ) ∈ S1 , (y1 , r2 , y2 ) ∈ S2 , and u1 = u − y2 . Show that the feedback connection Sfeedback has the transfer function −1 G1 (s) · (I + G2 (s)G1 (s)) at s, provided I + G2 (s)G1 (s) is invertible. u1

u −

y2

S1

S2

y1

y

u2

Figure 12.4: Feedback connection

12.4 Notes and references The idea for defining the transfer function via exponentially solutions is old, but has hardly been investigated for distributed parameter systems. [64] was the first paper where this approach has been used for infinite-dimensional systems. We note that in this paper a transfer function is called a characteristic function. One may find the exponential solution in Polderman and Willems [47], where all solutions of this type are called the exponential behavior. The formula for the transfer function, G(s) = C(sI −A)−1 B +D can also easily be derived using the Laplace transform, see e.g. [10]. However, via the Laplace transform approach the function is only defined in some right half-plane and not

170


on the whole resolvent set of A. In finite-dimensional spaces the transfer function is rational, and there are no mathematical difficulties. Therefore, the transfer function can easily be extended to ρ(A). The situation is different √ for infinitedimensional spaces, since transfer functions can contain terms like s, and for these functions it is less clear how to extend them. Positive realness is a well-studied property of transfer functions, see e.g. [1]. It first appeared in the study of electrical networks. Any (ideal) electrical network consisting of inductors, capacitors, and gyrators has a positive real transfer function, and even the converse holds, i.e., every (rational) positive real transfer function can be realized by such a network, see e.g. [2]. Our examples show that this property appears quite often and that positive realness is closely related to the energy balance.

Chapter 13

Well-posedness The concept of well-posedness can easily be explained by means of the abstract linear system Σ(A, B, C, D) introduced in Section 10.2, that is, for the set of equations x(t) ˙ = Ax(t) + Bu(t), y(t) = Cx(t) + Du(t),

x(0) = x0 ,

(13.1) (13.2)

where x is a X-valued function, u is an U -valued function and y is a Y -valued function. The spaces U , X and Y are assumed to be Hilbert spaces. Further, the operator A is assumed to be the infinitesimal generator of a C0 -semigroup, and B, C, and D are bounded linear operators. Under these assumptions the abstract differential equation (13.1)–(13.2) possesses for every u ∈ L1 ([0, τ ]; U ) a unique (mild) solution, see Theorem 10.2.1. Since L2 ([0, τ ]; U ) ⊂ L1 ([0, τ ]; U ) for τ < ∞, the same assertion holds for u ∈ L2 ([0, τ ]; U ). Existence of (mild) solutions for an arbitrary initial condition x0 ∈ X and an arbitrary input u ∈ L2 ([0, τ ]; U ), such that x is continuous and y ∈ L2 ([0, τ ]; Y ) is called well-posedness. Under the assumption that B, C, and D are bounded linear operators, the system (13.1)– (13.2) is well-posed if and only if A is the infinitesimal generator of a C0 -semigroup. The aim of this chapter is to extend this result to our class of port-Hamiltonian systems.

13.1 Well-posedness for boundary control systems In this section we turn to the definition of well-posedness. Although well-posedness can be defined in a quite general setup, see the notes and references section, we restrict ourselves to the class of boundary control systems as introduced in Chapter 11. We showed that boundary control systems in general cannot be written in the form (13.1)–(13.2) with B, C, and D bounded. However, we know that for every initial condition and every smooth input function there exists a (unique) B. Jacob and H.J. Zwart, Linear Port-Hamiltonian Systems on Infinite-dimensional Spaces, Operator Theory: Advances and Applications 223, DOI 10.1007/978-3-0348-0399-1_13, © Springer Basel 2012

171

172

Chapter 13. Well-posedness

mild solution of the state differential equation given by (11.9). Furthermore, from Theorem 11.2.1 follows that for smooth initial conditions and smooth inputs the output is well-defined. In the following example we see that the solution can often be extended to a larger set of initial conditions and input signals. Example 13.1.1. Consider the controlled transport equation on the interval [0, 1] with scalar control and observation on the boundary ∂x ∂x (ζ, t) = (ζ, t), ∂t ∂ζ u(t) = x(1, t), y(t) = x(0, t).

x(ζ, 0) = x0 (ζ),

ζ ∈ [0, 1],

(13.3) (13.4) (13.5)

From Example 11.1.4 follows that the mild solution of (13.3)–(13.4) is given by " x0 (ζ + t), ζ + t ≤ 1, x(ζ, t) = (13.6) u(ζ + t − 1), ζ + t > 1. For every t ≥ 0 the function x(·, t) is an element of X = L2 (0, 1), whenever u ∈ L2 (0, τ ) and x0 ∈ X. Furthermore, x(·, t) is a continuous function in t, i.e.,

x(·, t) − x(·, t + h) converges to zero when h converges to zero, see Exercise 6.3. Hence the mild solution (13.6) can be extended from controls in H 1 (0, τ ) to L2 (0, τ ). If x0 and u are smooth, then we clearly see that y(t) is well-defined for every t ≥ 0 and it is given by " x0 (t), 0 ≤ t ≤ 1, y(t) = (13.7) u(t − 1), t > 1. However, if x0 ∈ L2 (0, 1) and u ∈ L2 (0, τ ), then the expression (13.7) still implies that y is well-defined as an L2 -function. Summarizing, we can define a (mild) solution for (13.3)–(13.5) for all x0 ∈ X and all u ∈ L2 (0, τ ). This solution defines a continuous state trajectory in the state space, and an output trajectory which is square integrable on every compact time interval. Hence this system is well-posed. In the previous example we showed well-posedness for a transport equation with control and observation at the boundary. Next we formally define wellposedness for the boundary control systems introduced in Section 11.1 and 11.2. Thus the system is given by x(t) ˙ = Ax(t) Bx(t) = u(t), Cx(t) = y(t).

x(0) = x0 ,

(13.8) (13.9) (13.10)

As in Chapter 11 we need the following assumptions, see also Definition 11.1.1.

13.1. Well-posedness for boundary control systems

173

Assumption 13.1.2. The operators defining system (13.8)–(13.10) satisfy: 1. A : D(A) ⊂ X → X, B : D(B) ⊂ X → U , C : D(A) ⊂ X → Y are linear operators, D(A) ⊂ D(B), and X, U , Y are Hilbert spaces. 2. The operator A : D(A) → X with D(A) = D(A) ∩ ker(B) and Ax = Ax

for x ∈ D(A)

(13.11)

is the infinitesimal generator of a C0 -semigroup (T (t))t≥0 on X. 3. There exists an operator B ∈ L(U, X) such that for all u ∈ U we have Bu ∈ D(A), AB ∈ L(U, X) and BBu = u,

u ∈ U.

(13.12)

4. The operator C is bounded from the domain of A to Y . Here D(A) is equipped with the graph norm. We remark that Condition 4 was not required in Theorem 11.2.1. However, all the examples will satisfy this condition. Definition 13.1.3. Consider the system (13.8)–(13.10) satisfying Assumption 13.1.2. We call this system well-posed if there exist a τ > 0 and mτ ≥ 0 such that for all x0 ∈ D(A) and u ∈ C 2 ([0, τ ]; U ) with u(0) = Bx0 we have

x(τ ) 2X +

τ

y(t) 2 dt ≤ mτ 0

x0 2X +

u(t) 2 dt .

τ

(13.13)

0

In general it is not easy to show that a boundary control system is well-posed. However, there is a special class of systems for which well-posedness can be proved easily. This result is formulated next. Proposition 13.1.4. Assume that the boundary control system (13.8)–(13.10) satisfies Assumption 13.1.2. If every classical solution of the system satisfies d

x(t) 2 = u(t) 2 − y(t) 2 , dt

(13.14)

then the system is well-posed. Proof. If we integrate equation (13.14) from 0 to τ , we find that

x(τ ) 2 − x(0) 2 =

u(t) 2 dt − 0

which implies (13.13) with mτ = 1.

τ

τ

y(t) 2 dt, 0

174


In this special case the boundary control system satisfies (13.13) for every τ > 0. In Theorem 13.1.7 we show that this holds for every well-posed boundary control system. Furthermore, if a boundary control system is well-posed, then we can define a (mild) solution of (13.8)–(13.10) for all x0 ∈ X and u ∈ L2 ([0, τ ]; U ) such that t → x(t) is a continuous function in X and y is square integrable. To prove these statements we need the following lemmas for classical solutions of the system (13.8)–(13.10). Lemma 13.1.5. Assume that the boundary control system (13.8)–(13.9) satisfies conditions 1, 2, and 3 of Assumption 13.1.2, x0 ∈ X, and u ∈ C 1 ([0, τ ]; U ). Then the mild solution x of the boundary control system can written as t t x(t) = T (t)x0 + T (t − s)ABu(s) ds − A T (t − s)Bu(s) ds. (13.15) 0

0

Proof. For u ∈ C ([0, τ ]; U ) Corollary 10.1.4 implies t t T (t)Bu(0) + T (t − s)B u(s) ˙ ds − Bu(t) = A T (t − s)Bu(s) ds. 1

0

0

Combining this with the formula for the mild solution, see (11.9), we obtain (13.15). Formula (13.15) holds in particular for classical solutions of boundary control systems if x0 ∈ D(A), and u ∈ C 2 ([0, t1 ]; U ) with Bx0 = u(0). For the proof of Theorem 13.1.7 we need certain properties of these classical solutions. The proof is left as an exercise to the reader, see Exercise 13.5. Lemma 13.1.6. Consider the boundary control system (13.8)–(13.10) satisfying conditions 1, 2, and 3 of Assumption 13.1.2, and let t1 > 0. For x0 ∈ D(A), and u ∈ C 2 ([0, t1 ]; U ) with Bx0 = u(0) we denote the corresponding classical solution by x(·; x0 , u) and y(t; x0 , u), see Theorem 11.2.1, respectively. Then the following results hold: 1. The functions x(·; x0 , u) and y(·; x0 , u) depend linearly on the initial condition and on the input, i.e., for all x1 , x2 ∈ D(A) and u1 , u2 ∈ C 2 ([0, t1 ]; U ) with Bx1 = u1 (0), Bx2 = u2 (0) and α, β ∈ K there holds x(t; αx1 + βx2 , αu1 + βu2 ) = αx(t; x1 , u1 ) + βx(t; x2 , u2 ),

t ∈ [0, t1 ] (13.16)

and a similar property holds for y(t; x0 , u). 2. The system is causal, i.e., if x0 ∈ D(A), and u1 ∈ C 2 ([0, t1 ]; U ) with Bx0 = u1 (0) and if u2 ∈ C 2 ([0, t1 ]; U ) satisfies u1 (t) = u2 (t) for 0 ≤ t ≤ τ < t1 , then x(t; x0 , u1 ) = x(t; x0 , u2 ),

t ∈ [0, τ ],

(13.17)

y(t; x0 , u1 ) = y(t; x0 , u2 ),

t ∈ [0, τ ].

(13.18)


175

3. For x0 ∈ D(A), u ∈ C 2 ([0, t1 ]; U ) with Bx0 = u(0) there holds x(τ1 + τ2 ; x0 , u) = x(τ1 ; x(τ2 ), u(· + τ2 )), y(τ1 + τ2 ; x0 , u) = y(τ1 ; x(τ2 ), u(· + τ2 ))

(13.19) (13.20)

x(τ2 ) = x(τ2 ; x0 , u),

(13.21)

where

with τ1 , τ2 > 0, and τ1 + τ2 ≤ t1 . The following theorem shows that if (13.13) holds for some τ > 0, then (13.13) holds for all τ > 0. However, note that the constant mτ depends on τ . Theorem 13.1.7. If the boundary control system (13.8)–(13.10) satisfying Assumption 13.1.2 is well-posed, then for all τ > 0 there exists a constant mτ > 0 such that (13.13) holds. Proof. Step 1. Using Theorem 11.2.1 and Lemma 13.1.6, for every t > 0 we can define the linear operator S(t) : D(S(t)) ⊂ X ⊕ L2 ([0, t]; U ) → X ⊕ L2 ([0, t]; Y ) by x0 x(t; x0 , u) S(t) = , (13.22) y(t; x0 , u) u with 3 4 D(S(t)) = [ xu0 ] ∈ X ⊕L2 ([0, t]; U ) | x0 ∈ D(A), u ∈ C 2 ([0, t]; U ), Bx0 = u(0) . (13.23) Here x is the classical solution of the boundary control system and y is the corresponding output. For [ fx ] ∈ X ⊕ L2 ([0, t]; Z) with Z a Hilbert space, we define the norm as t x 2 2 =

x

+

f (t) 2Z dt. X f 2 X,L

0

As the boundary control system is well-posed there exist τ0 > 0 and mτ0 ≥ 0 such that 2 x(τ0 ; x0 , u) 2 x0 2 S(τ0 ) x0 = ≤ mτ 0 u X,L2 y(τ0 ; x0 , u) X,L2 u X,L2 for [ xu0 ] ∈ D(S(τ0 )). Since S(τ0 ) is densely defined, there exists a unique bounded √ extension S(τ0 ) with norm less than or equal to mτ0 . Equation (13.22) implies that the extension S(τ0 ) can be written as S1 (τ0 ) S1 (τ0 ) S(τ0 ) = , with S(τ0 ) = . (13.24) S2 (τ0 ) S2 (τ0 )

176


Step 2. In this step we prove that for λ ∈ ρ(A) and [ xu0 ] ∈ X ⊕ L2 ([0, t]; U ) the following holds: x −1 (13.25) (λI − A) S1 (τ0 ) 0 = T (τ )(λI − A)−1 x0 u τ + T (τ − s) (λI − A)−1 AB − A(λI − A)−1 B u(s) ds. 0

Using the definition of S1 (τ0 ), see (13.24), and Lemma 13.1.5, (13.25) holds for [ xu0 ] ∈ D(S(τ0 )). Since D(S(τ0 )) is dense in X ⊕L2 ([0, τ0 ]; U ), and since both sides of equation (13.25) are well-defined for x0 ∈ X and u ∈ L2 ([0, τ0 ]; U ) we conclude that this equality holds for all initial conditions and all input functions. For functions f , g ∈ L2 , we define " (f ♦ g)(t) = τ

f (t), g(t − τ ),

t < τ, t > τ.

(13.26)

Step 3. Next we show that S1 (τ ) has a bounded extension for τ ∈ (0, τ0 ). Let τ ∈ (0, τ0 ) be arbitrary. We define the operator S1 (τ ) by 5 6 0 x0 S1 (τ ) (13.27) := T (τ )x0 + S1 (τ0 ) 0 ♦ u . u τ −τ 0

Note that S1 (τ ) is defined by (13.27) and that we will only show in the sequel that S1 (τ ) is indeed the bounded extension of S1 (τ ). Clearly S1 (τ ) is a bounded operator from X ⊕L2 ([0, τ ]; U ) to X. It remains to show that S1 (τ ) is an extension of S1 (τ ). For λ ∈ ρ(A) and [ xu0 ] ∈ X ⊕ L2 ([0, τ ]; U ), (13.27) and (13.25) imply x −1 (λI − A) S1 (τ ) 0 u 5 6 0 −1 −1 = (λI − A) T (τ )x0 + (λI − A) S1 (τ0 ) 0 ♦ u τ0 −τ

−1

= (λI − A) T (τ )x0 τ0 + T (τ0 − s) (λI − A)−1 AB − A(λI − A)−1 B (0 ♦ u)(s) ds τ0 −τ

0 −1

= (λI − A) T (τ )x0 τ + T (τ − s) (λI − A)−1 AB − A(λI − A)−1 B u(s) ds.

(13.28)

0

Using Corollary 10.1.4, for [ xu0 ] ∈ D(S(τ )) the integral maps into D(A), and thus

13.1. Well-posedness for boundary control systems we obtain

177

x0 (λI − A) S1 (τ ) = (λI − A)−1 T (τ )x0 u τ −1 −1 + (λI − A) T (τ − s)ABu(s) ds − (λI − A) A −1

0

τ

T (τ − s)Bu(s) ds.

0

Multiplying from the left by (λI − A), we obtain that, for [ xu0 ] ∈ D(S(τ )), τ τ x S1 (τ ) 0 = T (τ )x0 + T (τ − s)ABu(s) ds − A T (τ − s)Bu(s) ds. (13.29) u 0 0 By Lemma 13.1.5, the right-hand side equals the classical state trajectory of the system (13.8)–(13.10). Thus by the definition of S(τ ), see (13.22), we obtain x0 x0 x0 S1 (τ ) = S1 (τ ) , ∈ D(S(τ )). u u u Thus S1 (τ ) is the unique bounded extension of S1 (τ ) to X ⊕ L2 ([0, τ ]; U ). Step 4. Let τ ∈ (0, τ0 ). We assume that [ ux01 ] , [ ux02 ] ∈ D(S(τ0 )) satisfy u1 (t) = u2 (t) for all t ∈ [0, τ ]. By Lemma 13.1.6, we obtain x0 x0 S2 (τ0 ) (t) = S2 (τ0 ) (t), t ∈ [0, τ ]. (13.30) u1 u2 Thus the bounded extension S2 (τ0 ) satisfies for all x0 ∈ X and u1 , u2 ∈ L2 ([0, τ0 ]; U ) with u1 (t) = u2 (t) a.e. for t ∈ [0, τ ], x0 x0 S2 (τ0 ) (t) = S2 (τ0 ) (t), for a.e. t ∈ [0, τ ]. (13.31) u1 u2 Step 5. Let [ xu0 ] ∈ D(S(τ )) and define uext ∈ C 2 ([0, τ0 ]; U ) as a C 2 -extension of u, 0 i.e., uext (t) = u(t) for t ∈ [0, τ ]. Then it is easy to see that [ uxext ] ∈ D(S(τ0 )). By Theorem 11.2.1 for t ∈ [0, τ ], x0 x0 S2 (τ ) (t) = S2 (τ0 ) (t). (13.32) u uext Using (13.31), (13.32), and the estimate (13.13) we obtain 5 62 2 2 x0 x x 0 0 S2 (τ ) = = S2 (τ0 ) u ♦ 0 S2 (τ0 ) u u 2 ext 2 2 L (0,τ ) L (0,τ ) τ L (0,τ ) 5 62 x0 ≤ S2 (τ0 ) u ♦ 0 2 τ L (0,τ0 ) ≤ mτ0 x0 2 + u ♦ 0 2L2 (0,τ0 ) τ 2 = mτ0 x0 + u 2L2 (0,τ ) .

178


Thus S2 (τ ) possesses a linear bounded extension from X ⊕ L2 ([0, τ ]; U ) to L ([0, τ ]; Y ). Combining this fact with the result of Step 3, we may conclude that (13.13) holds for τ with mτ = S(τ ) . Step 6. Next we show that (13.13) holds for τ > τ0 . We first assume that τ ∈ x0 (τ0 , 2τ0 ], and we write τ = τ0 + t1 with t1 ∈ (0, τ0 ]. For [ u ] ∈ D(S(τ )) it is easy x0 to see that u|[0,τ0 ] ∈ D(S(τ0 )). Using equations (13.19)–(13.21), the following equations hold: x0 x(τ0 ) S1 (τ ) = S1 (t1 ) , (13.33) u(τ0 + ·) u 5 6* ⎧) x0 ⎪ ⎪ ⎪ S (τ ) (t), t ≤ τ0 , ⎪ ⎨ 2 0 u x0 5 6* ) S2 (τ ) (t) = (13.34) u x(τ0 ) ⎪ ⎪ ⎪ (t), t ∈ (τ0 , τ ], S (t ) ⎪ ⎩ 2 1 u(τ0 + ·) 2

where x(τ0 ) = S1 (τ0 )

x0 u|[0,τ0 ]

.

(13.35)

Using these equations and the boundedness of the operators S1 (t), S2 (t), t ∈ (0, τ0 ], we may conclude that the operators S1 (τ ) and S2 (τ ) possess bounded extensions from X ⊕ L2 ([0, τ ]; U ) to X and to L2 ([0, τ ]; Y ), respectively. Therefore, (13.13) holds for τ with mτ = S(τ ) . The general case τ > 2τ0 follows by induction. In the proof of the previous theorem we have introduced the operators S(t) mapping the initial condition and input to the state at time t and the output on the time interval [0, t]. If the boundary control system is well-posed, then we may extend S(t) to a bounded linear operator from X ⊕ L2 ([0, t]; U ) to X ⊕ L2 ([0, t]; Y ) and we can decompose the operator S(t) as S11 (t) S12 (t) S(t) = . (13.36) S21 (t) S22 (t) Equation (13.27) implies that S11 (t) equals the semigroup at time t. We are now in the position to extend the definition of a mild solution, see Section 11.1. Definition 13.1.8. Let (13.8)–(13.10) be a well-posed boundary control system with input space U and output space Y and let S(t) be decomposed as in (13.36). For x ∈ X and u ∈ L2 ([0, τ ]; U ) the mild solution of (13.8)–(13.10) is given by x(t) = S11 (t)x0 + S12 (t)u, y = S21 (τ )x0 + S22 (τ )u.

t ∈ [0, τ ],

(13.37)


179

Remark 13.1.9. For a fixed τ > 0 it follows from the proof of Theorem 13.1.7 that there exists a constant mτ ≥ 0 such that for every t ∈ (0, τ ] we have S12 (t)u ≤ mτ u . Combining this observation with the fact that the classical solutions are continuous, it is easy to show that the state trajectory x of a mild solution of (13.8)–(13.10) is continuous. Note that the output is only square integrable. The mild state space solution defined in Definition 13.1.8 extends the mild solution as defined by (11.9). The fact that S11 (t)x0 + S12 (t)u equals (11.9) for x0 ∈ X and u ∈ H 1 ([0, τ ]; U ) can be shown as in Step 3 of the proof of Theorem 13.1.7. In Chapter 12 we showed that the boundary control system (13.8)–(13.10) possesses a transfer function. Using Theorem 12.1.3 we conclude that this function is defined on the resolvent set of A. Since A generates a C0 -semigroup, the resolvent set contains a right half-plane, see Proposition 5.2.4, which implies that the transfer function exists on some right-half plane. Furthermore, the transfer function is bounded as Re(s) → ∞ for a well-posed system. Lemma 13.1.10. If G is the transfer function of a well-posed system, then

G(s) ≤ for Re(s) ≥

log(mτ ) , 2τ

√ mτ

(13.38)

where τ and mτ are the constants of equation (13.13).

τ) Proof. Choose s ∈ C such that mτ ≤ |e2sτ |, i.e., Re(s) ≥ log(m . For this 2τ fixed s ∈ C and an arbitrary u0 ∈ U we consider the exponential solution (u0 est , x0 est , G(s)u0 est )t≥0 . Inequality (13.13) implies that τ τ 2sτ 2 2 st 2 2 2 st 2 |e | x0 X +

G(s)u0 |e | dt ≤ mτ x0 X +

u0 |e | dt 0 0 τ ≤ |e2sτ | x0 2X + mτ

u0 2 |est |2 dt.

0

This inequality shows that G(s)u0 2 ≤ mτ u0 2 . Since u0 is arbitrary, we conclude that (13.38) holds. Although the transfer function is bounded on some right half-plane, this does not imply that the transfer function converges along the real axis. If it does converge, then the boundary control system is called regular. Definition 13.1.11. Let G be the transfer function of a well-posed boundary control system. The boundary control system is called regular if lims∈R,s→∞ G(s) exists. If the boundary control system is regular, then the feed-through term D is defined as D = lims∈R,s→∞ G(s). Well-posed regular boundary control systems are closed under a class of output feedback operators and bounded perturbations of the generator. These results were first obtained by G. Weiss. The proofs of these result lie outside the scope of this chapter, and we refer the interested reader to [60] or [51].

180

Chapter 13. Well-posedness v

u

−

y

S

y

F

Figure 13.1: The closed loop system Theorem 13.1.12. Assume that the boundary system (13.8)–(13.10) satisfying Assumption 13.1.2, denoted by S, is well-posed. We denote the corresponding transfer function by G. Let F be a bounded linear operator from Y to U and assume that the inverse of I + G(s)F exists and is bounded for s in some right half-plane. Then the closed loop system as depicted in Figure 13.1 is again well-posed, that is, the boundary control system x(t) ˙ = Ax(t)

(13.39)

v(t) = (B + F C)x(t) y(t) = Cx(t)

(13.40) (13.41)

satisfies Assumption 13.1.2 and it is well-posed. If the boundary control system S is regular, then the closed loop system is regular as well. The operator F is called a feedback operator. Remark 13.1.13. A feedback operator can be used to stabilize a system. More precisely, a feedback operator changes the operator B, that is, if we start with an unstable system, then via a clever choice of the feedback operator F we may stabilize the system. For instance, in Example 9.2.1 the boundary condition (9.28) can be interpreted as a feedback relation. Namely, the input (force at the righthand side) is equal to the feedback, −k, times the output (velocity at the righthand side). Lemma 13.1.14. Consider the boundary system (13.8)–(13.10) satisfying the conditions of Assumption 13.1.2, and let Q be a bounded operator on X, i.e., Q ∈ L(X). This system is well-posed if and only if the system x(t) ˙ = Ax(t) + Qx(t)

(13.42)

with inputs and outputs given by (13.9)–(13.10) is well-posed. Let G denote the transfer function of (13.8)–(13.10) and GQ the transfer function of (13.42) with (13.9)–(13.10). Then lim

G(s) =

lim

G(s) =

s∈R,s→∞

lim

GQ (s),

(13.43)

lim

GQ (s).

(13.44)

s∈R,s→∞

and Re(s)→∞

Re(s)→∞

13.2. Well-posedness for port-Hamiltonian systems

181

13.2 Well-posedness for port-Hamiltonian systems We now turn our attention to the class of port-Hamiltonian systems with boundary control and boundary observation as introduced in Sections 11.3, that is, to systems of the form ∂x ∂ (ζ, t) = P1 (H(ζ)x(ζ, t)) + P0 (H(ζ)x(ζ, t)), (13.45) ∂t ∂ζ f (t) u(t) = WB,1 ∂ , (13.46) e∂ (t) f (t) 0 = WB,2 ∂ , (13.47) e∂ (t) f (t) y(t) = WC ∂ . (13.48) e∂ (t) W We assume that P1 , H and WB := WB,1 satisfy Assumption 11.3.1. Thus in B,2 particular, for a.e. ζ ∈ [a, b], H(ζ) is a self-adjoint n × n matrix satisfying 0 < mI ≤ H(ζ) ≤ M I. Furthermore, WB :=

WB,1 WB,2

is a full rank matrix of size

n × 2n. We assume that WB,1 is a m × 2n matrix . Recall that the state space is given by the Hilbert space X = L2 ([a, b]; Kn ) with the inner product 1 b f, g X = f (ζ)∗ H(ζ)g(ζ)d ζ. (13.49) 2 a 1

The following lemma follows easily from Theorems 11.3.2 and 11.3.5, see also Exercise 11.4. Lemma 13.2.1. Let τ be a positive real number. Assume that the operator A := ∂ P1 ∂ζ H + P0 H with domain ' ( ∂,0 D(A) = x0 ∈ X | Hx0 ∈ H 1 ([a, b]; Kn ), WB fe∂,0 =0 is the infinitesimal generator of a C0 -semigroup on X. Then the system (13.45)– (13.47) is a boundary control system satisfying Assumption 13.1.2. n 2 m In particular, for every Hx0 ∈ H 1([a, b]; K ) and every u ∈ C ([0, τ ]; K ) ∂,0 , and 0 = WB,2 fe∂,0 , there exists a unique classical sowith u(0) = WB,1 fe∂,0 ∂,0 lution of (13.45)–(13.47) on [0, τ ]. Furthermore, the output (13.48) is well-defined and y is continuous on [0, τ ]. By Lemma 13.2.1, if A generates a C0 -semigroup on X, then the system possesses classical solutions for smooth inputs and initial conditions. Well-posedness implies that there exist solutions for every initial condition and every square integrable input. Now we are in the position to formulate our main result. 1 Note that m has two meanings: It is used as a lower-bound for H and as the dimension of our input space.

182


Theorem 13.2.2. Consider the port-Hamiltonian system (13.45)–(13.48) and assume that the conditions of Assumption 11.3.1 are satisfied. Furthermore, we assume that 1. The multiplication operator P1 H can be written as P1 H(ζ) = S −1 (ζ)Δ(ζ)S(ζ),

ζ ∈ [a, b],

(13.50)

where Δ is a diagonal matrix-valued function, S is a matrix-valued function and both Δ and S are continuously differentiable on [a, b]. WB,1 2. rank WB,2 = n + rank (WC ). WC

Let X be the Hilbert space L2 ([a, b]; Kn ) with inner product (13.49). If the operator A corresponding to the homogeneous p.d.e., i.e., u ≡ 0, generates a C0 -semigroup on X, then the system (13.45)–(13.48) is regular, and thus in particular well-posed. Furthermore, we have that limRe(s)→∞ G(s) = lims→∞,s∈R G(s). The proof of Theorem 13.2.2 will be given in Sections 13.3 and 13.4. Assumption 11.3.1 and the Assumptions 1 and 2 of Theorem 13.2.2 can be easily checked. Thus it basically remains to check that A generates a C0 -semigroup. Note that this operator equals the operator A defined in Lemma 13.2.1. From Chapter 7 we know that if WB ΣWB∗ ≥ 0 and P0 = −P0∗ , then A generates a (contraction) semigroup, and thus in this situation the system is well-posed. In particular, we obtain a mild solution for all square integrable inputs. Remark 13.2.3. 1. Assumption 11.3.1 is very standard, and is assumed to be satisfied for all our port-Hamiltonian systems until now. 2. Note that we do not have any assumption on P0 . In fact the term P0 H may be replaced by any bounded operator on X. This result will be proved in Section 13.4. 3. Assumption 1 of Theorem 13.2.2 concerning the multiplication operator P1 H is not very strong, and will almost always be satisfied if H is continuously differentiable. Note that Δ contains the eigenvalues of P1 H, whereas S −1 contains the eigenvectors. 4. The last condition implies that we are not measuring quantities that are set to zero, or chosen to be an input. This condition is not important for the proof, and will normally follow from correct modeling. As mentioned above, Assumption 1 of Theorem 13.2.2 is quite weak. We illustrate this by means of an example, the wave equation studied in Example 7.1.1.


183

Example 13.2.4. From Example 7.1.1 together with equation (7.5) follows that the port-Hamiltonian formulation of the wave equation is given by ∂ ∂t

x1 (ζ, t) x2 (ζ, t)

=

0 1 1 0

∂ ∂ζ

1 ρ(ζ)

0

0 T (ζ)

x1 (ζ, t) x2 (ζ, t)

,

(13.51)

∂w where x1 = ρ ∂w ∂t is the momentum and x2 = ∂ζ is the strain. Hence we have that 1 0 0 1 ρ(ζ) , H(ζ) = P1 = . 1 0 0 T (ζ)

Being physical constants, the Young’s modulus T and the mass density ρ are positive. Hence P1 and H satisfy the first two conditions of Assumption 11.3.1. Under the assumption that T and ρ are continuously differentiable, we show that (13.50) holds. The eigenvalues of P1 H(ζ) =

0 1 ρ(ζ)

T (ζ) 0

(13.52)

(ζ) are given by ±γ(ζ) with γ(ζ) = Tρ(ζ) . The corresponding eigenvectors are given by γ(ζ) −γ(ζ) and . (13.53) 1 1 ρ(ζ)

ρ(ζ)

Hence P1 H = S

−1

ΔS =

γ 1 ρ

−γ 1 ρ

γ 0

0 −γ

1 2γ 1 − 2γ

ρ 2 ρ 2

,

(13.54)

where we have omitted the dependence on ζ. This shows that the assumptions of Theorem 13.2.2 are satisfied. In Example 13.2.4 the matrix P1 H possesses only real eigenvalues and the number of positive and negative eigenvalues is independent of ζ. This can be shown to hold in general. Lemma 13.2.5. Let P1 and H satisfy the conditions of Theorem 13.2.2. Then Δ can be written as Λ 0 Δ= , (13.55) 0 Θ where Λ is a diagonal real matrix-valued function, with (strictly) positive functions on the diagonal, and Θ is a diagonal real matrix-valued function, with (strictly) negative functions on the diagonal.

184


Proof. Since H(ζ) > mI for every ζ ∈ [a, b], the square root of H(ζ) exists for 1 1 every ζ ∈ [a, b]. By Sylvester’s law of inertia, the inertia of H(ζ) 2 P1 H(ζ) 2 equals 1 1 the inertia of P1 . This implies that the inertia of H(ζ) 2 P1 H(ζ) 2 is independent of 1 1 ζ. Thus, since P1 is invertible, zero is not an eigenvalue of H(ζ) 2 P1 H(ζ) 2 and the 1 1 number of negative eigenvalues of H(ζ) 2 P1 H(ζ) 2 equals the number of negative eigenvalues of P1 . A similar statement holds for the positive eigenvalues. 1 1 A simple calculation shows that the eigenvalues of H(ζ) 2 P1 H(ζ) 2 are equal to the eigenvalues of P1 H(ζ). Concluding, for every ζ ∈ [a, b] zero is not an eigenvalue of P1 H(ζ), and the number of negative and positive eigenvalues of P1 H(ζ) is independent of ζ. By regrouping the eigenvalues, we obtain the requested representation (13.55). The proof of Theorem 13.2.2 will be given in Sections 13.3 and 13.4. As can be guessed from the condition (13.50) we will diagonalize P1 H. Therefore it is essential to know the systems properties for the case that P1 H is diagonal. In this situation it suffices to understand the systems behavior for a positive and a negative element. This is the subject of the following two lemmas. Lemma 13.2.6. Let λ be a positive continuous function on the interval [a, b]. Then the port-Hamiltonian system ∂w ∂ (ζ, t) = (λ(ζ)w(ζ, t)) , ∂t ∂ζ u(t) = λ(b)w(b, t),

w(ζ, 0) = w0 (ζ),

y(t) = λ(a)w(a, t),

ζ ∈ [a, b],

(13.56) (13.57) (13.58)

is a regular system on the state space L2 (a, b) and the corresponding transfer function G satisfies lim G(s) = 0. (13.59) Re(s)→∞

Proof. It is easy to see that the system (13.56)–(13.58) is a very simple version of the general Port-Hamiltonian system (13.45)–(13.48) with P1 = 1, H = λ, √ √ √ √ 2 2 2 WB = [ 2 2 ] and WC = [− 2 22 ]. Since WB ΣWB∗ = 1 > 0, we conclude that the operator A corresponding to the homogeneous p.d.e., i.e., u ≡ 0, generates a C0 -semigroup on L2 (a, b). Thus Theorem 11.3.5 implies that (13.56)–(13.58) has a well-defined solution provided the initial condition and the input are smooth. For this class, the following balance equation holds, see (12.29) d b w(ζ, t)λ(ζ)w(ζ, t)dζ = |λ(b)w(b, t)|2 − |λ(a)w(a, t)|2 dt a = |u(t)|2 − |y(t)|2 . By Proposition 13.1.4 we conclude that the system is well-posed on the energy b space. Since λ is strictly positive, we have that the energy norm a |w(ζ, t)|2 λ(ζ)dζ 2 is equivalent to the L (a, b)-norm, and so the system is also well-posed on L2 (a, b).


185

Example 12.2.2 shows that the transfer function of (13.56)–(13.58) is given by G(s) = e−p(b)s with p(b) > 0. Thus the system is regular and property (13.59) follows immediately. For a p.d.e. with negative coefficient, we obtain a similar result. Lemma 13.2.7. Let θ be a negative continuous function on the interval [a, b]. We consider the system ∂w ∂ (ζ, t) = (θ(ζ)w(ζ, t)) , ∂t ∂ζ

w(ζ, 0) = w0 (ζ),

ζ ∈ [a, b].

(13.60)

The value at a we choose as input u(t) = θ(a)w(a, t)

(13.61)

and as output we choose the value on the other end y(t) = θ(b)w(b, t).

(13.62)

The port-Hamiltonian system (13.60)–(13.62) is regular on the state space L2 (a, b). Its transfer function is given by G(s) = en(b)s , where

n(ζ) =

ζ

θ(z)−1 dz,

(13.63)

ζ ∈ [a, b].

(13.64)

a

The transfer function satisfies lim

G(s) = 0.

(13.65)

Re(s)→∞

Lemma 13.2.7 will be proved in Exercise 13.7. In Theorem 13.2.2 we asserted that under some weak conditions every portHamiltonian system is well-posed provided the corresponding homogeneous equation generates a strongly continuous semigroup. In Sections 13.3 and 13.4 we prove this assertion. However, we like to explain the main ideas of the proof by means of an example. We choose the wave equation of Example 13.2.4. Since S is invertible, well-posedness will not change if we perform a basis transformation, x ˜ = Sx, see Exercise 13.3. After this basis transformation, the p.d.e. (13.51) reads ∂x ˜ ∂ dS −1 (ζ) (ζ, t) = (Δ˜ x) (ζ, t) + S(ζ) Δ(ζ)˜ x(ζ, t) ∂t ∂ζ dζ ∂ dS −1 (ζ) γ(ζ)˜ x1 (ζ, t) = Δ(ζ)˜ x(ζ, t). + S(ζ) x2 (ζ, t) ∂ζ −γ(ζ)˜ dζ

(13.66)

186


Thus via a basis transformation we obtain a set of simple p.d.e.’s, just two simple −1 x(ζ, t). transport equations, but they are corrupted by the term S(ζ) dS dζ (ζ) Δ(ζ)˜ We first assume that this term is not present, and so we study the well-posedness of the collection of transport equations ∂ ∂ x ˜1 (ζ, t) γ(ζ)˜ x1 (ζ, t) = . (13.67) ˜2 (ζ, t) x2 (ζ, t) ∂t x ∂ζ −γ(ζ)˜ Although it seems that these two p.d.e.’s are uncoupled, they are coupled via the boundary conditions. Ignoring the coupling via the boundary condition, we can apply Lemmas 13.2.6 and 13.2.7 directly. In Section 13.3 we investigate when the p.d.e. (13.67) with control and observation at the boundary is well-posed. In Section 13.4 we return to the original p.d.e., and apply Lemma 13.1.14 to show that ignoring bounded terms, like we did in (13.67) does not influence the wellposedness of the system. Since a basis transformation does not effect it either, we have proved Theorem 13.2.2 for the wave equation.

13.3 P1 H diagonal In this section, we prove Theorem 13.2.2 in the situation that P1 H is diagonal, i.e., when S = I. Thus we consider the following diagonal port-Hamiltonian system ∂ ∂ x+ (ζ, t) Λ(ζ) 0 x+ (ζ, t) = , (13.68) x− (ζ, t) 0 Θ(ζ) ∂t x− (ζ, t) ∂ζ where for every ζ ∈ [a, b], Λ(ζ) is a diagonal (real) matrix, with positive numbers on the diagonal, and Θ(ζ) is a diagonal (real) matrix, with negative numbers on the diagonal. we assume that Λ and Θ are continuously differentiable Furthermore, 0 is an n×n-matrix. We note that (13.68) is a port-Hamiltonian and that Λ(ζ) 0 Θ(ζ) Λ 0

I 0

system with H := 0 −Θ and P1 := 0 −I . The following boundary control and observation are of interest Λ(b)x+ (b, t) us (t) := , (13.69) Θ(a)x− (a, t) ys (t) :=

Λ(a)x+ (a, t) Θ(b)x− (b, t)

.

(13.70)

Theorem 13.3.1. Consider the p.d.e. (13.68) with boundary control us and boundary observation ys as defined in (13.69) and (13.70), respectively. • The system defined by (13.68)–(13.70) is well-posed and regular. Furthermore, the corresponding transfer function Gs converges to zero for Re(s) → ∞.

13.3. P1 H diagonal

187

• We equip the port-Hamiltonian system (13.68) with a new set of inputs and outputs. The new input u(t) is of the form u(t) = Kus (t) + Qys (t), where K and Q are two square n × n-matrices with K new output is of the form

(13.71)

Q of rank n. The

y(t) = O1 us (t) + O2 ys (t),

(13.72)

where O1 and O2 are k × n-matrices. Concerning system (13.68) with input u and output y, we have the following results: 1. If K is invertible, then the system (13.68), (13.71), and (13.72) is well-posed and regular. Furthermore, the corresponding transfer function converges to O1 K −1 for Re(s) → ∞ 2. If K is not invertible, then the operator AK defined as ∂ g+ (ζ) Λ(ζ) 0 g+ (ζ) AK = g− (ζ) 0 Θ(ζ) g− (ζ) ∂ζ with domain + D(AK ) =

g+ (ζ) g− (ζ)

(13.73)

∈ H ([a, b], K ) (13.74) , Λ(a)g+ (a) Λ(b)g+ (b) +Q =0 K Θ(a)g− (a) Θ(b)g− (b)

1

n

does not generate a C0 -semigroup on L2 ([a, b]; Kn ). Note that part 2 implies that the homogeneous p.d.e. does not have a welldefined solution, when K is not invertible. Proof. The first part is a direct consequence of Lemma 13.2.6 and 13.2.7 by noticing that the system (13.68)–(13.70) is built out of copies of the system (13.56)– (13.58) and the system (13.60)–(13.62). Furthermore, these sub-systems do not interact with each other. Thus in particular, the corresponding transfer function Gs (s) is diagonal. For the proof of the first part of the second assertion, with K invertible, we rewrite the new input, as us (t) = K −1 u(t) − K −1 Qys (t), t ≥ 0. This can be seen as a feedback interconnection on the system (13.68)–(13.70), as is depicted in Figure 13.2. Note that the system (13.68)–(13.70) is denoted by Ss . The system contains one feedback loop with feedback K −1 Q. By Theorem 13.1.12, we have that, if I + Gs (s)K −1 Q is invertible for some complex s and if this inverse exists and is bounded on a right half-plane, then the closed loop system is well-posed. Since limRe(s)→∞ Gs (s) = 0, this holds for every K −1 and Q. Thus under the

188

Chapter 13. Well-posedness y

O1

u

K −1

us

−

ys

Ss

O2

K −1 Q

Figure 13.2: The system (13.68) with input (13.71) and output (13.72)

assumption that K is invertible, the system (13.68) with input and output given by (13.71) and (13.72) is well-posed. The regularity follows easily. By regarding the loops in Figure 13.2, we see that the feed-through term is given by O1 K −1 . It remains to show that AK does not generate a C0 -semigroup if K is not invertible. Since v ∈ Kn such that v ∗ K = 0.

K is singular, there exists a non-zero ∗ ∗ Since K Q has full rank, we obtain that q := v Q = 0. This implies that at least one of the components of the vector q is unequal to zero. For the sake of the argument, we assume that this holds for the first component. Assume that AK is the infinitesimal generator of a C0 -semigroup. This implies that for every x0 ∈ D(AK ) the abstract differential equation x(t) ˙ = AK x(t),

x(0) = x0

(13.75)

solution, i.e., for every x0 ∈ D(AK ) there exists a function x(ζ, t) := has a classical x+ (ζ,t) x− (ζ,t) which is a classical solution to the p.d.e. (13.68) with initial condition x(·, 0) = x0 , and satisfies for all t ≥ 0 the boundary condition K

Λ(b)x+ (b, t) Θ(a)x− (a, t)

+Q

Λ(a)x+ (a, t) Θ(b)x− (b, t)

= 0.

Using the vectors v and q, the function x satisfies 0=q

∗

Λ(a)x+ (a, t) Θ(b)x− (b, t)

,

t ≥ 0.

(13.76)

Now we construct an initial condition in D(AK ), for which this equality does not hold. Note that we have chosen the first component of q unequal to zero. The initial condition x0 is chosen to have all components zero except for the first one. For this first component we choose an arbitrary function in H 1 (a, b) which is zero at a and b, but nonzero everywhere else on the open set (a, b). Clearly this initial condition is an element of the domain of AK . Now we solve (13.68).

13.4. Proof of Theorem 13.2.2

189

Standard p.d.e. theory implies that the solution of (13.68) can be written as x+,m (ζ, t) = f+,m (pm (ζ) + t)λm (ζ)−1 , −1

x−, (ζ, t) = f−, (n (ζ) + t)θ (ζ)

(13.77)

,

(13.78)

where λm and θ are the m-th and the -th diagonal element of Λ and Θ, reζ ζ spectively. Furthermore, pm (ζ) = a λm (ζ)−1 dζ, n (ζ) = a θ (ζ)−1 dζ, see also Exercises 13.6 and 13.7. In particular, we have pm (a) = n (a) = 0. The functions f+,m , f−, need to be determined from the boundary and initial conditions. Using the initial condition, we obtain f+,m (pm (ζ)) = λm (ζ)x0,+,m (ζ) and f−, (n (ζ)) = θ (ζ)x0,−, (ζ). Since pm > 0, and n < 0, the initial condition determines f+,m on a (small) positive interval, and f−, on a small negative interval. By our choice of the initial condition, we find that −1 f+,1 (ζ) = λ1 (p−1 1 (ζ))x0,+,1 (p1 (ζ)), f+,m (ζ) = 0,

f−, (ζ) = 0,

ζ ∈ [0, p1 (b)), ζ ∈ [0, pm (b)),

m ≥ 2,

ζ ∈ [n (b), 0),

≥ 1.

(13.79)

The solution x(ζ, t) must also satisfy (13.76), thus for every t > 0 we have ⎤ ⎡ f+,1 (t) ⎥ ⎢ .. ⎥ ⎢ . ⎥ 0 = q∗ ⎢ (13.80) ⎢ f−,1 (n1 (b) + t) ⎥ . ⎦ ⎣ .. . Combining this with (13.79), we find 0 = q1 f+,1 (p1 (ζ)) = q1 x0,+,1 (ζ)λ1 (ζ) on some interval [a, β]. Since q1 and λ1 are unequal to zero, we find that x0,+,1 must be zero on some interval. This is in contradiction to our choice of the initial condition. Thus AK cannot be the infinitesimal generator of a C0 -semigroup.

13.4 Proof of Theorem 13.2.2 In this section we use the results of the previous section to prove Theorem 13.2.2. By Assumption 1 of Theorem 13.2.2, the matrices P1 and H satisfy the equation (13.50): P1 H(ζ) = S −1 (ζ)Δ(ζ)S(ζ). We introduce the new state vector x ˜(ζ, t) = S(ζ)x(ζ, t),

ζ ∈ [a, b].

(13.81)

190


Under this basis transformation, the p.d.e. (13.45) becomes ∂x ˜ ∂ dS −1 (ζ) (ζ, t) = (Δ˜ x) (ζ, t) + S(ζ) Δ(ζ)˜ x(ζ, t) ∂t ∂ζ dζ + S(ζ)P0 (ζ)S(ζ)−1 x ˜(ζ, t), x˜0 (ζ) := x ˜(ζ, 0) = S(ζ)x0 (ζ). (13.82) The equations (13.46)–(13.48) imply the existence of matrices M11 , M12 , M21 , ˜ 11 , M ˜ 12 , M ˜ 21 , M ˜ 22 , C1 , C2 , C˜1 and C˜2 such that M22 , M 0 = M11 P1−1 S −1 (b)Δ(b)˜ x(b, t) + M12 P1−1 S −1 (a)Δ(a)˜ x(a, t) ˜ 11 Δ(b)˜ ˜ 12 Δ(a)˜ =M x(b, t) + M x(a, t), M21 P1−1 S −1 (b)Δ(b)˜ x(b, t)

u(t) =

+

˜ 21 Δ(b)˜ ˜ 22 Δ(a)˜ =M x(b, t) + M x(a, t), C1 P1−1 S −1 (b)Δ(b)˜ x(b, t)

y(t) =

(13.84)

C2 P1−1 S −1 (a)Δ(a)˜ x(a, t)

+ ˜ ˜ = C1 Δ(b)˜ x(b, t) + C2 Δ(a)˜ x(a, t).

˜ = We introduce M

˜ j1 M

˜ j2 M

and C˜ =

C˜1

Since the matrix

˜ 11 M ˜ 12 M ˜ 21 M ˜ 22 M

=

Mj2

Mj1

C˜2

and C˜ =

=

C1

C˜1

(13.85)

C˜2 . We have

P1−1 S(b)−1 0

C2

P1−1 S(b)−1 0 0 P1−1 S(a)−1

(13.83)

M22 P1−1 S −1 (a)Δ(a)˜ x(a, t)

0 P1−1 S(a)−1

P1−1 S(b)−1 0

,

0 P1−1 S(a)−1

j = 1, 2 .

has full rank, the rank conditions in The˜ and C. ˜ orem 13.2.2 imply similar rank conditions for M Using Lemma 13.1.14, we only have to prove the result for the p.d.e. ∂ ∂x ˜ (ζ, t) = (Δ˜ x) (ζ, t) ∂t ∂ζ

(13.86)

with boundary conditions, inputs, and outputs as described in (13.83)–(13.85). If condition (13.83) is not present, then Theorem 13.3.1 implies that the above system is well-posed and regular if and only if the homogeneous p.d.e. generates a C0 -semigroup on L2 ([a, b]; Kn ). Since the state transformation (13.81) defines a bounded mapping on L2 ([a, b]; Kn ), we have proved Theorem 13.2.2 provided there is no condition of the form (13.47). Thus it remains to prove Theorem 13.2.2 in the case that we have set part of the boundary conditions to zero. Or equivalently, to prove that the system (13.83)–(13.86) is well-posed and regular if and only if the homogeneous p.d.e. generates a C0 -semigroup.

13.5. Well-posedness of the vibrating string

191

We replace (13.83) by ˜ 11 Δ(b)˜ ˜ 12 Δ(a)˜ v(t) = M x(b, t) + M x(a, t),

(13.87)

where we regard v as a new input. Hence we have the system (13.86) with the new extended input ˜ 11 ˜ 12 v(t) M M = x(t, b) + x(t, a) (13.88) ˜ 21 Δ(b)˜ ˜ 22 Δ(a)˜ u(t) M M and the output (13.85). Thus we obtain a system without a condition of the form (13.83). For the new system we know that it is well-posed and regular if and only if the homogeneous equation generates a C0 -semigroup. Assume that the system (13.86), (13.88) and (13.85) is well-posed, then we may choose any (locally) square integrable input. In particular, we may choose v ≡ 0. Thus the system (13.83)–(13.86) is well-posed and regular as well. Assume next that the p.d.e. with the extended input in (13.88) set to zero does not generate a C0 -semigroup. Since this gives the same homogeneous p.d.e. as (13.86) with (13.83) and u in (13.84) set to zero, we know that this p.d.e. does not generate a C0 -semigroup either. This finally proves Theorem 13.2.2.

13.5 Well-posedness of the vibrating string In this section we illustrate the usefulness of Theorem 13.2.2 by applying it to the vibrating string of Example 7.1.1. By Example 13.2.4 for the vibrating string we have 1 0 0 1 . P1 = , H(ζ) = ρ(ζ) 1 0 0 T (ζ) We want to illustrate the theory and proofs derived in the previous sections, and therefore we do not directly check if for a (to-be-given) set of boundary conditions that the semigroup condition is satisfied. Instead of that, we rewrite the system in its diagonal form, and check the conditions of Theorem 13.3.1. As we have seen in Section 13.4, the proof of Theorem 13.2.2 follows after a basis transformation directly from Theorem 13.3.1. Hence we first diagonalize P1 H. Although all the results hardly change, we assume for simplicity of notation that Young’s modulus T and the mass density ρ are constant. Being physical constants, they are naturally positive. From equation (13.54) we know that the operator P1 H is diagonalizable: 1 ρ γ −γ γ 0 −1 2γ 2 P1 H = S ΔS = 1 , (13.89) 1 ρ 1 0 −γ − 2γ ρ ρ 2 where γ is positive and satisfies γ 2 =

T ρ.

192

Chapter 13. Well-posedness Hence the state transformation under which the p.d.e. becomes diagonal is 1 x˜ = 2

1 γ − γ1

ρ ρ

x.

Since we assumed that γ > 0, the functions x˜1 and x˜2 correspond to x+ and x− in equation (13.68), respectively and Λ, Θ to γ and −γ, respectively. Hence the input and output us and ys defined for the diagonal system (13.68) by the equations (13.69)–(13.70) are expressed in the original coordinates by 1 x1 (b, t) + γρx2 (b, t) us (t) = , (13.90) 2 x1 (a, t) − γρx2 (a, t) 1 x1 (a, t) + γρx2 (a, t) ys (t) = . (13.91) 2 x1 (b, t) − γρx2 (b, t) This pair of boundary input and output variables consists of complementary linear combinations of the momentum x1 and the strain x2 at the boundaries: however they lack an obvious physical interpretation. One could consider another choice of boundary input and outputs, for instance the velocity and the strain at the boundary points and choose as input x1 ρ (b, t) u1 (t) = (13.92) x2 (a, t) and as output

y1 (t) =

x1 (a, t) ρ

.

x2 (b, t)

(13.93)

We may apply Theorem 13.3.1 to check whether this system is well-posed, and to find the feed-through. Expressing the input-output pair (u1 , y1 ) in terms of (us , ys ) gives 6 5 6 5 1 1 0 0 ρ ρ ys (t), u1 (t) = us (t) + √1 (13.94) 0 √−1 0 Tρ Tρ 6 5 6 5 1 1 0 0 ρ ρ y1 (t) = √1 us (t) + ys (t). (13.95) 0 √−1 0 Tρ Tρ Hence

5 K= 5 O1 =

1 ρ

6

0

0

√−1 Tρ

0

1 ρ

√1 Tρ

0

5 Q=

0 √1 Tρ

5

6 ,

O2 =

1 ρ

0

1 ρ

6

0 0 √−1 Tρ

,

(13.96) 6 .

(13.97)

13.6. Exercises

193

Since K is invertible, the system with the input-output pair (u1 , y1 ) is well-posed 0 −γ and regular, and the feed-through term is given by O1 K −1 = γ1 0 . Since the states are defined as x1 = ρ ∂w and x2 = ∂t

∂w

, the control and ∂t (t,b) and y1 (t) = ∂w (t,a)

∂ζ ∂w

observation are easily formulated using w. Namely, u1 (t) = ∂ζ ∂w (t,a) ∂t , respectively. Hence we observe the velocity and strain at opposite ends. ∂w (t,b) ∂ζ

Next we show that if we control the velocity and strain at the same end, this does not give a well-posed system. The control and observation are given by ∂w x1 (b, t) (b, t) ∂t ρ u2 (t) = ∂w = (13.98) x2 (b, t) ∂ζ (b, t) and as output

y2 (t) =

∂w ∂t (a, t) ∂w (a, t) ∂ζ

=

x1 ρ (a, t)

.

x2 (a, t)

(13.99)

It is easy to see that u2 (t) =

1 ρ 1 γρ

0 0

us (t) +

1 0 ρ 1 0 − γρ

ys (t).

(13.100)

Clearly the matrix in front of us is not invertible, and hence we conclude by Theorem 13.3.1 that the wave equation with the homogeneous boundary conditions u2 = 0 does not generate a C0 -semigroup. Hence this system is not well-posed.

13.6 Exercises 13.1. In this exercise we show that a homogeneous port-Hamiltonian system needs not to generate a contraction semigroup. Consider the transport equation ∂x ∂x (ζ, t) = (ζ, t), ∂t ∂ζ x(1, t) = 2x(0, t).

x(ζ, 0) = x0 (ζ),

ζ ∈ [0, 1],

(13.101) (13.102)

(a) Show that for an initial condition x0 which is continuously differentiable and satisfies x0 (1) = 2x0 (0) and x 0 (1) = 2x 0 (0) the (classical) solution of (13.101)–(13.102) is given by x(ζ, t) = 2n x0 (τ ),

(13.103)

where ζ + t = n + τ , τ ∈ [0, 1], n ∈ N ∪ {0}. (b) Show that the mapping x0 → x(·, t) with x(ζ, t) given by (13.103) defines a C0 -semigroup on L2 (0, 1).

194

Chapter 13. Well-posedness (c) Conclude that (13.103) is the mild solution of (13.101)–(13.102) for any initial condition x0 ∈ L2 (0, 1).

13.2. Let X be a Hilbert space with norm, · X and let · n be an equivalent norm, i.e., there exists α, β > 0 such that for all x ∈ X there holds α x X ≤

x n ≤ β x X . Show that if boundary control system (13.8)–(13.10) satisfying Assumption 13.1.2 is well-posed with respect to the state space norm · X , then the system is also well-posed with respect to the state space norm · n . 13.3. Consider the boundary control system (13.8)–(13.10) satisfying Assumption 13.1.2. Let S be a be linear operator in L(X) which is boundedly invertible. Introduce the new state variable x ˜ = Sx. (a) Rewrite the boundary control system (13.8)–(13.10) as a boundary control system in the state variable x ˜. Furthermore, show that this new boundary control system satisfies Assumption 13.1.2. (b) Show that the transformed boundary control system is again well-posed when the original boundary control system is well-posed. (c) Prove that both systems have the same transfer function. 13.4. Assume the the boundary control system (13.8)–(13.10) is well-posed and define (13.104) ynew := O1 u(t) + O2 y(t), where O1 and O2 are bounded linear operators. Show the port-Hamiltonian systems remains well-posed if we replace (13.10) by (13.104). 13.5. Use Theorem 11.2.1 to prove Lemma 13.1.6. 13.6. In this exercise we prove some further results concerning the system (13.56)– (13.58). (a) Show that the system (13.56)–(13.58) is a port-Hamiltonian system of the form (13.45)–(13.48). (b) Show that the solution of (13.56)–(13.57) is given by w(ζ, t) = f (p(ζ) + t)λ(ζ)−1 ,

(13.105)

where p(ζ) =

ζ

λ(z)−1 dz,

ζ ∈ [a, b],

(13.106)

ζ ∈ [a, b],

(13.107)

t > 0.

(13.108)

a

f (p(ζ)) = λ(ζ)w0 (ζ), f (p(b) + t) = u(t), 13.7. In this exercise we prove Lemma 13.2.7.


195

(a) Show that the solution of (13.60)–(13.61) is given as w(ζ, t) = f (n(ζ) + t)θ(ζ)−1 , where

n(ζ) =

ζ

θ(z)−1 dz,

(13.109)

(13.110)

a

f (n(ζ)) = θ(ζ)w0 (ζ), z ∈ [a, b], f (t) = u(t), t > 0.

(13.111) (13.112)

(b) Use part a to prove Lemma 13.2.7.

13.7 Notes and references We have defined well-posedness for boundary control systems. However, it is also possible to define well-posedness more abstractly. In order to do this we start S11 (t) S12 (t) with a bounded linear operator-valued function (S(t))t≥0 = , S21 (t) S22 (t) t≥0

satisfying properties (13.27), (13.33)–(13.35) and (S11 (t))t≥0 is a C0 -semigroup, see the book of Staffans [51]. Our class of port-Hamiltonian systems is most likely the only class of p.d.e.’s which allows for such a simple characterization of well-posedness. For a general p.d.e. it is usually very hard to prove well-posedness. The first condition of Theorem 13.2.2 implies that P1 H is diagonalizable via a continuously differentiable basis transformation. Since S(ζ) in equation (13.50) contains the eigenvectors, we need the eigenvectors to be continuously differentiable. Example II.5.3 of Kato [30] gives a symmetric matrix-valued function which is infinitely continuously differentiable, but the eigenvectors are not even once continuously differentiable. If P1 H is a C 1 -function such that for all ζ ∈ [a, b], the eigenvalues P1 H(ζ) are simple, then the eigenvalues and eigenvectors can be chosen continuously differentiable, see Kato [30, Theorem II.5.4]. In our proof of Theorem 13.2.2 we only use that Δ is continuous. However, when P1 H is continuously differentiable, then so are its eigenvalues. This follows 1 1 from the fact that the eigenvalues of P1 H and H 2 P1 H 2 are the same, and Theorem II.6.8 of [30]. From the proof of Theorem 13.2.2 follows that the operator A corresponding to the homogeneous equation generates a C0 -semigroup if and only if the matrix K is invertible, see Theorem 13.3.1. Since the matrix K is obtained after a basis transformation, and depends on the negative and positive eigenvalues of P1 H, it is not easy to rewrite this condition in terms of the matrix WB . A semigroup can be extended to a C0 -group, if for every initial condition the homogeneous p.d.e. has a solution for negative time. Using once more the proof of Theorem 13.3.1, the operator A of Theorem 13.2.2 generates a C0 -group if and only if K and Q are invertible matrices.

196


Theorem 13.1.12 was first proved by Weiss [60]. Moreover, we note that the content of this chapter is based on the paper by Zwart et al., [65].

Appendix A

Integration and Hardy Spaces A.1 Integration theory In this section, we wish to extend the ideas of Lebesgue integration of complexvalued functions to vector-valued and operator-valued functions, which take their values in a separable Hilbert space X or in the Banach space L(X1 , X2 ), where X1 , X2 are separable Hilbert spaces. By L(X1 , X2 ) we denote the class of all bounded, linear operators from X1 to X2 . As main references, we have used Arendt, Batty, Hieber and Neubrander [3], Diestel and Uhl [12], Dunford and Schwartz [13], and Hille and Phillips [24]. Throughout this section, we use the notation Ω for a closed subset of R, and (Ω, B, λ) for the measure space with the Lebesgue measure λ and (Lebesgue) measurable subsets B. It is possible to develop a Lebesgue integration theory based on various measurability concepts. Definition A.1.1. Let W be a Banach space. A function f : Ω → W is called simple if there exist w1 , w2 , . . . , wn ∈ W and E1 , E2 , . . . , En ∈ B such that f = ! n i=1 wi ½Ei , where ½Ei (t) = 1 if t ∈ Ei and 0 otherwise. Let X1 , X2 be two separable Hilbert spaces, and let F : Ω → L(X1 , X2 ) and f : Ω → X1 . 1. F is uniformly (Lebesgue) measurable if there exists a sequence of simple functions Fn : Ω → L(X1 , X2 ) such that lim Fn (t) = F (t) for almost all t ∈ Ω.

n→∞

2. f is strongly (Lebesgue) measurable if there exists a sequence of simple functions fn : Ω → X1 such that lim fn (t) = f (t) for almost all t ∈ Ω.

n→∞

F is strongly measurable if F x1 is strongly measurable for every x1 ∈ X1 . B. Jacob and H.J. Zwart, Linear Port-Hamiltonian Systems on Infinite-dimensional Spaces, Operator Theory: Advances and Applications 223, DOI 10.1007/978-3-0348-0399-1, © Springer Basel 2012

197

198

Appendix A. Integration and Hardy Spaces

3. f is weakly (Lebesgue) measurable if f, x1 is measurable for every x1 ∈ X1 . F is weakly measurable if F x1 is weakly measurable for every x1 ∈ X1 . It is easy to see that uniform measurability implies strong measurability, which implies weak measurability. For our separable Hilbert spaces X1 and X2 , the concepts weak and strong measurability coalesce. Lemma A.1.2. For the case that X1 and X2 are separable Hilbert spaces the concepts of weak and strong measurability in Definition A.1.1 coincide. Proof. See Hille and Phillips [24, theorem 3.5.3] or Yosida [61, theorem in Section V.4]. We often consider the inner product of two weakly measurable functions. Lemma A.1.3. Let X be a separable Hilbert space, and let f1 , f2 : Ω → X be two weakly measurable functions. The complex-valued function f1 (·), f2 (·) defined by the inner product of these functions is a measurable function. Proof. This follows directly from Lemma A.1.2 and Definition A.1.1.

Lemma A.1.4. Let f : Ω → X1 and F : Ω → L(X1 , X2 ) be weakly measurable, then f and F are measurable, scalar-valued functions. Proof. The first assertion follows directly from the previous lemma, since f = $ f (·), f (·) . Since X1 and X2 are separable, there exist countable dense subsets Z1 and Z2 of X1 and X2 , respectively. Since these sets are dense, we obtain

F (t) =

sup z1 ∈Z1 , z1 =0 z2 ∈Z2 , z2 =0

|F (t)z1 , z2 | ,

z1

z2

t ∈ Ω.

1 ,z2 | By assumption the functions |F z(t)z are measurable, and since a (countable) 1 z2 supremum of measurable functions is measurable, we conclude that F (·) is measurable.

The notion of the Lebesgue integral follows naturally from the measurability concepts given in Definition A.1.1. Definition A.1.5. Suppose that (Ω, B, λ) is the Lebesgue measure space and that E ∈ B. 1. Let W !nbe a Banach space and let f : Ω → W be a simple function given by f = i=1 wi ½Ei , where the Ei are disjoint. We define f to ! be Lebesgue inten grable over E if f is Lebesgue integrable over E, that is, i=1 wi λ(Ei ∩ E) < ∞, where λ(·) denotes the Lebesgue measure of the set and we follow the usual! convention that 0 · ∞ = 0. The Lebesgue integral of f over E is n given by i=1 wi λ(Ei ∩ E) and will be denoted by E f (t)dt.

A.1. Integration theory

199

2. Let X1 and X2 be two separable Hilbert spaces. The uniformly measurable function F : Ω → L(X1 , X2 ) is Lebesgue integrable over E if there exists a sequence of simple integrable functions Fn converging almost everywhere to F and such that lim

F (t) − Fn (t) L(X1 ,X2 ) dt = 0. n→∞

E

We define the Lebesgue integral by F (t)dt = lim Fn (t)dt. n→∞

E

E

3. Let X be a separable Hilbert space. The strongly measurable function f : Ω → X is Lebesgue integrable over E if there exists a sequence of simple integrable functions fn converging almost everywhere to f and such that lim

f (t) − fn (t) X dt = 0. n→∞

E

We define the Lebesgue integral by f (t)dt = lim fn (t)dt. E

n→∞

E

The integrals in the above definition are also called Bochner integrals in the literature. If f : Ω → X is Lebesgue integrable over E, then E f (t)dt is an element of X. Similarly, if f : Ω → L(X1 , X2 ) is Lebesgue integrable over E, then F (t)dt ∈ L(X1 , X2 ). For functions from R to a separable Hilbert space X, there E is a simple criterion to test whether a function is Lebesgue integrable. Lemma A.1.6. Let f : Ω → X, where X is a separable Hilbert space. f is Lebesgue integrable over E ∈ B if and only if the function x, f (·) is measurable for every x ∈ X and E f (t) dt < ∞. Proof. See Arendt, Batty, Hieber and Neubrander [3, Theorem 1.1.4], or Hille and Phillips [24, theorem 3.7.4], noting that weak and strong measurability are the same for separable Hilbert spaces (Lemma A.1.2). Proposition A.1.7. Let X be a separable Hilbert space, let f : [a, b] → X be Lebesgue t integrable and F (t) := a f (s) ds for t ∈ [a, b]. Then F is differentiable a.e. and (t) F = f a.e., that is, for almost all t ∈ [a, b] we have limh→0 F (t+h)−F = f (t). h Proof. See Arendt, Batty, Hieber and Neubrander [3, Theorem 1.2.2].

In the case of operator-valued functions F : Ω → L(X1 , X2 ), where X1 and X2 are separable Hilbert spaces, we need to distinguish between the Lebesgue integral E F (t) dt for the case that F is uniformly (Lebesgue) measurable and the Lebesgue integral E F (t)x dt for the case that F is only strongly (Lebesgue) measurable.

200


Example A.1.8. Let (T (t))t≥0 be a C0 -semigroup on a separable Hilbert space X. Since (T (t))t≥0 is strongly continuous, it is strongly measurable. In fact, Hille and Phillips [24, theorem 10.2.1] show that the C0 -semigroup is uniformly measurable if and only if it is uniformly continuous. Now the only uniformly continuous semigroups are those whose infinitesimal generator (see Definition 5.2.1) is a bounded operator, Hille and Phillips [24, theorem 9.4.2], and so (T (t))t≥0 will only be 1

strongly measurable in general. Thus 0 T (t)x dt is a well-defined Lebesgue inte1 gral for any x ∈ X, but 0 T (t) dt is in general not well-defined. Next we study whether

τ 0

T (τ − s)F (s)ds has a meaning.

Example A.1.9. Let (T (t))t≥0 be a C0 -semigroup on a separable Hilbert space X, and let F : [0, ∞) → L(U, X) be weakly measurable, U is a Hilbert space, and

F ∈ L1 (0, τ ). Since (T ∗ (t))t≥0 is also a C0 -semigroup, T ∗ (·)x is continuous, see [10, Theorem 2.2.6], and so strongly measurable. Furthermore, by definition, we have that F (·)u is weakly measurable. Hence Lemma A.1.3 shows that x, T (τ − ·)F (·)u = T ∗ (τ − ·)x, F (·)u is measurable for all x ∈ X, u ∈ U . So from Lemma τ A.1.6 we have that for each u ∈ U 0 T (τ − s)F (s)uds is a well-defined Lebesgue τ integral. However, 0 T (τ − s)F (s)ds need not be a well-defined Lebesgue integral, since the integrand will not to be uniformly measurable in general. If an operator-valued function is not uniformly measurable, but only weakly measurable, there is still the possibility to define the so-called Pettis integral. Definition A.1.10. Let X1 and X2 be separable Hilbert spaces and let F : Ω → L(X1 , X2 ). If for all x1 ∈ X1 and x2 ∈ X2 we have that the function x2 , F (·)x1 ∈ L1 (Ω), then we say that F is Pettis integrable. Furthermore, for all E ∈ B, we call F (t)dt defined by E

F (t)dtx1 :=

x2 , E

x2 , F (t)x1 dt

(A.1)

E

the Pettis integral of F over E and E.

E

F (t)x1 dt the Pettis integral of F (·)x1 over

As for the Lebesgue integral we have that the Pettis integral E F (t)dt is an element of L(X1 , X2 ). It also has the usual properties such as linearity (αF1 (t) + βF2 (t)) dt = α F1 (t)dt + β F2 (t)dt. (A.2) E

E

E

From the definition of the Pettis integral, we see that a weakly measurable function F : Ω → L(X1 , X2 ) is Pettis integrable if and only if |x2 , F (t)x1 |dt < ∞. (A.3) E

A.1. Integration theory

201

In particular, if this weakly measurable function satisfies E F (t) dt < ∞, then the condition (A.3) is satisfied and so it is Pettis integrable. Furthermore, it is easy to see that if F is an integrable simple function, then the Pettis integral equals the Lebesgue integral. From the definition of the Lebesgue integral, it follows easily that if the Lebesgue integral of a function exists, then the Pettis integral also exists, and they are equal. In the following examples, we re-examine Examples A.1.8 and A.1.9, which we considered as Lebesgue integrals. Example A.1.11. We recall from Example A.1.8 that the C0 -semigroup (T (t))t≥0 on the separable Hilbert space X is in general only strongly measurable and so 1 1 while 0 T (t)xdt exists as a Lebesgue integral 0 T (t)dt does in general not. We show next that it does exist as a Pettis integral. Since (T (t))t≥0 is strongly continuous, we have that x1 , T (t)x2 is measurable for every x1 , x2 ∈ X. From Theorem 1 5.1.5 we have that 0 T (t) dt < ∞. Thus by Definition A.1.10 the Pettis integral 1 0 T (t)dt is well-defined. If the infinitesimal generator A of (T (t))t≥0 is invertible, then using Theorem 5.2.2 we can even calculate this Pettis integral to obtain 1

T (t)dt = A−1 T (1) − A−1 .

0 τ

Example A.1.12. From Example A.1.9 we recall that 0 T (τ − s)F (s)ds is in general not a well-defined Lebesgue integral. We already showed that x, T (τ − ·) F (·)u is Lebesgue measurable for all x ∈ X, u ∈ U . Furthermore, we see that

τ

T (τ − s)F (s) ds ≤ Mω e 0

τ

F (s) ds < ∞.

ωτ 0

τ

τ

So by Definition A.1.10 the integrals 0 T (τ − s)F (s)ds and 0 T (τ − s)F (s)uds are well-defined as Pettis integrals, where only the latter one is well-defined as a Lebesgue integral. Most of the integrals we use in this lecture notes satisfy the conditions in Lemma A.1.6. τ

Example A.1.13. Consider 0 T (τ − s)Bu(s)ds, where (T (t))t≥0 is a C0 -semigroup on a separable Hilbert space X, B ∈ L(U, X), U is a separable Hilbert space and u ∈ L1 ([0, τ ]; U ) (see Definition A.1.14). Then, as in Example A.1.12, τ x, T (τ − ·)Bu(·) is measurable for all x ∈ X and 0 T (τ − s)Bu(s) ds ≤ τ ωτ Mω e B 0 u(s) ds < ∞. So by Lemma A.1.6, the integral is well-defined as a Pettis or as a Lebesgue integral. To avoid confusion between the Pettis and Lebesgue integrals we introduce the following notation.

202


Definition A.1.14. Let X1 , X2 , and X be separable Hilbert spaces, and let Ω be a closed subset of R. We define the following spaces: P (Ω; L(X1 , X2 )) := {F : Ω → L(X1 , X2 ) | x2 , F (·)x1 is measurable for every x1 ∈ X1 and x2 ∈ X2 }. P (Ω; L(X1 , X2 )) := {F ∈ P (Ω; L(X1 , X2 )) | F p := 1/p p

F (t) L(X1 ,X2 ) dt < ∞}; 1 ≤ p < ∞. p

Ω

P ∞ (Ω; L(X1 , X2 )) := {F ∈ P (Ω; L(X1 , X2 )) | F ∞ := ess supΩ F (t) L(X1 ,X2 ) < ∞}. L(Ω; X) := {f : Ω → X | x, f (·) is measurable for all x ∈ X}. 1/p p p

f (t) X dt < ∞}; 1 ≤ p < ∞. L (Ω; X) := {f ∈ L(Ω; X) | f p := Ω

L∞ (Ω; X) := {f ∈ L(Ω; X) | f ∞ := ess supΩ f (t) X < ∞}. The reason for using the ”L” notation is that these integrals are also defined in the Lebesgue sense. For example, if (T (t))t≥0 is a strongly continuous semigroup, then T (·)x ∈ Lp ([0, τ ]; X) for all x ∈ X, but we only have that T (·) ∈ P p ([0, τ ]; L(X)) instead of the Lebesgue space Lp ([0, τ ]; L(X)) (see Example A.1.11). We remark that if X1 and X2 are finite-dimensional, then L(X1 , X2 ) is also finite-dimensional, and so L∞ (Ω; L(X1 , X2 )) is well-defined as a Lebesgue space (see Lemma A.1.6) and equals P ∞ (Ω; L(X1 , X2 )). Lemma A.1.15. If we do not distinguish between two functions that differ on a set of measure zero, then the spaces P p (Ω; L(X1 , X2 )), P ∞ (Ω; L(X1 , X2 )), Lp (Ω; X), and L∞ (Ω; X) are Banach spaces. Furthermore, L2 (Ω; X) is a Hilbert space with inner product h, f = h(t), f (t) X dt. (A.4) Ω

Proof. See Thomas [53] or [52]. The completeness property of Lp is also shown in theorem III.6.6 of Dunford and Schwartz [13]. In Section 3.5 of Balakrishnan [5] it is shown that L2 (Ω, X) is a Hilbert space.

A.2 The Hardy spaces In this section, we consider some special classes of functions that are holomorphic on the open half-plane C+ 0 := {s ∈ C | Re(s) > 0}.

A.2. The Hardy spaces

203

Good general references for this section are Kawata [31] and Helson [23] for the scalar case and Thomas [53] and Rosenblum and Rovnyak [49] for the vectorvalued case. As in finite dimensions, we define holomorphicity of a complex-valued function as differentiability. Definition A.2.1. Let X1 and X2 be Hilbert spaces, and let F : Υ → L(X1 , X2 ), where Υ is a domain in C. Then F is holomorphic on Υ if F is weakly differentiable on Υ, i.e., for all x1 ∈ X1 , x2 ∈ X2 , the function F (·)x1 , x2 is differentiable. Example A.2.2. Let A be a closed linear operator on the Hilbert space X. Define F : ρ(A) → L(X) by F (λ) = (λI − A)−1 . We shall prove that F is holomorphic on ρ(A). The resolvent equation (5.12) implies ((λ + h)I − A)−1 x1 , x2 − (λI − A)−1 x1 , x2 = −h(λI − A)−1 ((λ + h)I − A)−1 x1 , x2 . Since F is continuous, this implies that F is weakly differentiable with −(λI − A)−2 . Thus the resolvent operator is holomorphic.

dF dλ (λ)

=

Our definition can be seen as “weak holomorphic”. We remark that uniform and weak holomorphicity are equivalent. Next we define special classes of holomorphic functions. Definition A.2.3. For a Banach space W and a separable Hilbert space X we define the following Hardy spaces: ' ( → W | G is holomorphic and sup

G(s)

< ∞ ; H∞ (W) := G : C+ Re(s)>0 0 + H2 (X) :=

f : C+ 0 → X | f is holomorphic and , ∞ 1 2 2

f 2 = sup(

f (ζ + iω) dω) < ∞ . ζ>0 2π −∞

(A.5)

When the Banach space W or the Hilbert space X equals C, we shall use the notation H∞ and H2 for H∞ (C) and H2 (C), respectively. In most of the literature, Hardy spaces on the disc are usually treated; see, for example, Rosenblum and Rovnyak [49]. Lemma A.2.4. If W is a Banach space, then H∞ (W) from Definition A.2.3 is a Banach space under the H∞ -norm

G ∞ :=

sup G(s) W .

(A.6)

Re(s)>0

Proof. See Theorem D of Rosenblum and Rovnyak [49, section 4.7]. We now collect several important results in the following lemma.

204


Lemma A.2.5. The following are important properties of H∞ (L(U, Y )), where U, Y are separable Hilbert spaces: 1. For every F ∈ H∞ (L(U, Y )) there exists a unique function F˜ ∈ P ∞ ((−i∞, i∞); L(U, Y )) such that lim F (x + iω)u = F˜ (iω)u x↓0

for all u ∈ U and almost all ω ∈ R

(i.e., F ∈ H∞ (L(U, Y )) has a well-defined extension to the boundary); 2. The mapping F → F˜ is linear, injective and norm preserving, i.e., sup F (s) L(U,Y ) = ess supω∈R F˜ (iω) L(U,Y )

Re(s)>0

(consequently, we can identify F ∈ H∞ (L(U, Y )) with its boundary function F˜ ∈ P ∞ ((−i∞, i∞); L(U, Y )) and we can regard H∞ (L(U, Y )) as a closed subspace of the Banach space P ∞ ((−i∞, i∞); L(U, Y ))); 3. Identifying F with F˜ , the following holds: sup F (s) L(U,Y ) = ess supω∈R F (iω) L(U,Y ) < ∞.

Re(s)≥0

Proof. See theorems A of sections 4.6 and 4.7 of Rosenblum and Rovnyak [49]. We remark that Rosenblum and Rovnyak [49] use the notation L∞ for P ∞ . In general, the boundary function F˜ will not have the property that F˜ is uniformly measurable in the L(U, Y ) topology; see Rosenblum and Rovnyak [49, exercise 1 of chapter 4] or Thomas [53]. Lemma A.2.6. H2 (X) is a Banach space under the H2 -norm defined by (A.5), and the following important properties hold: 1. For each f ∈ H2 (X) there exists a unique function f˜ ∈ L2 ((−i∞, i∞); X) such that lim f (x + iω) = f˜(iω) for almost all ω ∈ R x↓0

and lim f (x + ·) − f˜(·) L2 ((−i∞,i∞);X) = 0; x↓0

2. The mapping f → f˜ is linear, injective, and f 22 = it is norm preserving

1 2π

∞ −∞

f˜(iω) 2 dω, i.e.,

(consequently, we identify the function f ∈ H2 (X) with its boundary function f˜ ∈ L2 ((−i∞, i∞);X) and regard H2 (X) as a closed subspace of L2 ((−i∞, i∞);X));


205

2 3. Denote by C+ α the subset {s ∈ C | Re(s) > α}. For any f ∈ H (X) and any α > 0 we have that

f (s) = 0 (A.7) lim sup ρ→∞

s∈C+ α ; |s|>ρ

(sometimes the terminology f (s) → 0 as |s| → ∞ in C+ α is used). Proof. Parts 1 and 2. The proof for the scalar case as given by Kawata [31, theorem 6.5.1]. Since this theorem holds for vector-valued function as well, the proof of parts a and b is similar to that for the scalar case. Part 3. See Hille and Phillips [24, theorem 6.4.2]. We remark that in general part 3 is not true for α = 0. ¿From this lemma we deduce the following result. Corollary A.2.7. If X is a separable Hilbert space, then H2 (X) is a Hilbert space under the inner product 1 f, g := 2π

∞ f (iω), g(iω) dω. −∞

H2 (X) is a very special Hilbert space, as is apparent from the following lemma and the Paley-Wiener theorem. Lemma A.2.8. Let X be a separable Hilbert space and let f ∈ H2 (X) be different from the zero function. Then f is nonzero almost everywhere on the imaginary axis. Proof. Suppose that there is a subset V of the imaginary axis with positive measure such that f is zero on this set. Then for every x ∈ X, we have that f, x ∈ H2 and it is zero on V . This implies that ∞ | log(f (iω), x )| dω = ∞. 1 + ω2 −∞ By Theorem 6.6.1 of Kawata [31] this can only happen if f, x is the zero function. Since x ∈ X was arbitrary, this would imply that f = 0. This is in contradiction to our assumption, and so the set V cannot have positive measure. Theorem A.2.9 (Paley-Wiener Theorem). If X is a separable Hilbert space, then under the Laplace transform L2 ([0, ∞); X) is isomorphic to H2 (X) and it preserves the inner products. Proof. See Thomas [53].

The following theorem gives a characterization of bounded operators between frequency-domain spaces.

206


Theorem A.2.10. Suppose that U and Y are separable Hilbert spaces. 1. If F ∈ P ∞ ((−i∞, i∞); L(U, Y )) and u ∈ L2 ((−i∞, i∞); U ), then F u ∈ L2 ((−i∞, i∞); Y ). Moreover, the multiplication map ΛF : u → F u defines a bounded linear operator from L2 ((−i∞, i∞); U ) to L2 ((−i∞, i∞); Y ), and

ΛF u L2 ((−i∞,i∞);Y ) ≤ F ∞ u L2((−i∞,i∞);U) , where · ∞ denotes the norm on P ∞ ((−i∞, i∞); L(U, Y )). In fact,

ΛF = sup u=0

ΛF u L2 ((−i∞,i∞);Y ) = F ∞ .

u L2 ((−i∞,i∞);U)

2. If F ∈ H∞ (L(U, Y )) and u ∈ H2 (U ), then F u ∈ H2 (Y ). Moreover, the multiplication map ΛF : u → F u defines a bounded linear operator from H2 (U ) to H2 (Y ), and

ΛF u H2 (Y ) ≤ F ∞ u H2 (U) , where · ∞ denotes the norm on H∞ (L(U, Y )). In fact,

ΛF = sup u=0

ΛF u H2 (Y ) = F ∞ .

u H2 (U)

3. F ∈ P ∞ ((−i∞, i∞); L(U, Y )) is in H∞ (L(U, Y )) if and only if ΛF H2 (U ) ⊂ H2 (Y ). Proof. Part 1. See Thomas [53]. Part 2. It is easy to show that for F ∈ H∞ (L(U, Y )) the first inequality holds. So 2

ΛF ≤ F ∞ . To prove the other inequality, let λ ∈ C+ 0 , y0 ∈ Y and f ∈ H (U ). Consider 8 8 9 9 y0 ∗ y0 f, ΛF = ΛF f, · + λ H2 (U) · + λ H2 (Y ) 9 ∞8 1 y0 = F (iω)f (iω), dω 2π −∞ iω + λ Y ∞ 1 −1 = F (iω)f (iω), y0 Y dω 2π −∞ iω − λ = F (λ)f (λ), y0 Y

by Cauchy’s Theorem

∗

= f (λ), F (λ) y0 U ∞ 1 −1 = f (iω), F (λ)∗ y0 U dω 2π −∞ iω − λ using Cauchy’s Theorem again


207 = f, F (λ)∗

y0 2 . · + λ H (U)

Since the above equality holds for every f ∈ H2 (U ), we have that Λ∗F

y0 y0 = F (λ)∗ . ·+λ ·+λ

This implies that Λ∗F ≥ F ∗ ∞ . Now the general property that F ∗ ∞ = F ∞ , concludes the proof. Part 3. See Thomas [53]. The proof of part 2 was communicated by George Weiss.

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Index A-invariant, 46 C 1 ([0, τ ]; X), 124 H 1 ([a, b]; Kn ), 85 L(Ω; X), 202 L(X), 52 L1loc ([0, ∞); Km ), 21 Lp (Ω; X), 202 L∞ (Ω; X), 202 P (Ω; L(X1 , X2 )), 202 P p (Ω; L(X1 , X2 )), 202 P ∞ (Ω; L(X1 , X2 )), 202 R(A, B), 28 TD (t)x0 , 132 Wt , 28 Σ(A, B), 130 Σ(A, B, C, D), 131 f˙, 1 df dt , 1 σ + , 135 σ − , 135 ½, 197 f (1) , 1 x(·; x0 , u), 174 y(t; x0 , u), 174 C+ 0 , 135 C− 0 , 135 K, 20 H2 , 203 H2 (X), 203 H∞ , 203 H∞ (W), 203 L(X1 , X2 ), 197

abstract differential equation, 124 A-invariant, 101 beam Timoshenko, 82 Bochner integrals, 199 boundary control systems, 144 boundary effort, 85, 86 boundary flow, 85, 86 boundary operator, 144 causal system, 174 Cayley-Hamilton Theorem, 28 classical solution, 20, 61 boundary control system, 144 on [0, ∞), 124 on [0, τ ], 124 coercive, 88 contraction semigroup, 65 controllability Gramian, 28 controllability matrix, 28 controllable, 27 controllable in time t1 , 29 C0 -semigroup, 53 growth bound, 55 measurable, 200 perturbed, 132 Datko’s lemma, 97 decay rate, 97 differential equation state, 13 dissipative, 68

B. Jacob and H.J. Zwart, Linear Port-Hamiltonian Systems on Infinite-dimensional Spaces, Operator Theory: Advances and Applications 223, DOI 10.1007/978-3-0348-0399-1, © Springer Basel 2012

215

216

Index

domain generator, 57

Kalman controllability decomposition, 35

eigenvalue multiplicity, 103, 133 order, 103 energy space, 81 exponential solution, 159 exponentially stable, 97 exponentially stabilizable, 133 exponentially stable, 39

Lebesgue integrable, 198, 199 Lebesgue integral, 198, 199 linear, first order port-Hamiltonian system, 81 Lumer-Phillips Theorem, 69 Lyapunov equation, 99

feed-through, 179 feedback, 133 feedback connection, 169 feedback operator, 133, 180 finite-dimensional systems, 13 group C0 , 73 strongly continuous, 73 unitary, 74 growth bound, 55 Hamiltonian, 23, 81 Hamiltonian density, 23 Hardy space, 203 heat conduction, 5 heat equation, 51 inhomogeneous, 129 Hermitian matrix, 82 Hille-Yosida Theorem, 66 holomorphic, 203 Hurwitz matrix, 40 infinitesimal generator, 57 input, 1 input space, 130 integral Bochner, 199 Lebesgue, 198 Pettis, 200 invariant A, 46, 101 T (t), 101

matrix Hermitian, 82 symmetric, 82 measurable of semigroups, 200 strong, 197 uniform, 197 weak, 198 mild solution, 21, 61, 126, 131 boundary control system, 146 well-posed system, 178 multiplicity, 103, 133 operator boundary, 144 order, 103 output, 1, 131 boundary control, 147 output space, 131 Paley-Wiener theorem, 205 parallel connection, 168 Pettis integrable, 200 Pettis integral, 200 pole placement problem, 40 port-Hamiltonian system, 23, 81, 84 positive real, 167 positive-definite matrix, 22 power, 80 power balance, 80 ran, 28 rank, 28 reachable, 28

Index regular, 179 resolvent operator, 59 resolvent set, 59 rk, 28 semigroup C0 , 53 contraction, 65 strongly continuous, 53 semigroup invariance, see T (t)-invariant series connection, 168 similar, 33 simple, 197 skew-adjoint, 75 solution classical, 20, 61, 124 boundary control system, 144 exponential, 159 mild, 21, 61, 126, 131 boundary control systems, 146 weak, 24, 127 spectral projection, 103 spectrum decomposition assumption at zero, 136 stability margin, 97 stabilizable, 40, see exponentially stabilizable exponentially, 133 stable, see exponentially stable exponentially, 39, 97 strongly, 108 state, 13, 53 state differential equation, 13 state space model, 13 state space representation, 13 state space system, 13 strongly (Lebesgue) measurable, 197 strongly measurable, 197 structure matrix, 23 Sturm-Liouville operator, 95 symmetric matrix, 82

217 system boundary control, 144 general, 159 Theorem Hille-Yosida, 66 Lumer-Phillips, 69 Theorem of Cayley-Hamilton, 28 Timoshenko beam, 82 transfer function, 159 regular, 179 transfer function at s, 159 transmission line, 93, 120 transport equation controlled, 143 T (t)-invariant, 101 uniformly (Lebesgue) measurable, 197 uniformly measurable, 197 unitary group, 74 variation of constant formula, 20 vibrating string, 4, 79 boundary control system, 153 wave equation, 4 weak solution, 24, 127, 128 weakly (Lebesgue) measurable, 198 weakly measurable, 198 well-posed, 173 well-posedness, 171

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